NCERT Solutions Class 12 maths Chapter-7 (Integrals)Exercise 7.4

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-7 (Integrals)Exercise 7.4 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.

We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Exercise 7.4

Q1. Integrate the function : $\frac{3 x^{2}}{x^{6} + 1}$

Answer. $\begin{aligned} Let x^{3} = t \\ ∴ 3 x^{2} d x = d t \\ \Rightarrow 3 x^{2} + 1 \\ \Rightarrow \int \frac{3 x^{2}}{x^{6} + 1} d x & = \int \frac{d t}{t^{2} + 1} \\ = \tan^{- 1} (t) + C = \tan^{- 1} (x^{3}) + C \end{aligned}$

Q2. Integrate the function : $\frac{1}{\sqrt{1 + 4 x^{2}}}$

Answer. Let 2x = t ∴ 2dx = dt $\begin{aligned} \Rightarrow \int \frac{1}{\sqrt{1 + 4 x^{2}}} d x & = \frac{1}{2} \int \frac{d t}{\sqrt{1 + t^{2}}} \\ = \frac{1}{2} [\log | t + \sqrt{t^{2} + 1} |] + C \end{aligned} \begin{array}{l} [\int \frac{1}{\sqrt{x^{2} + a^{2}}} d t = \log | x + \sqrt{x^{2} + a^{2}} |] \end{array}$ $= \frac{1}{2} \log | 2 x + \sqrt{4 x^{2} + 1} | + C$

Q3. Integrate the function : $\frac{1}{\sqrt{(2 - x)^{2} + 1}}$

\sqrt{Answer. Let 2 − x = t ⇒ −dx = dt⇒ ∫ 1 √ (2 − x) 2 + 1 d x = − ∫ 1 √ t 2 + 1 d t = − log ∣ ∣ t + √ t 2 + 1 ∣ ∣ + C [∫ 1 √ x 2 + a 2 d t = log ∣ ∣ x + √ x 2 + a 2 ∣ ∣] \begin{aligned} \Rightarrow \int \frac{1}{\sqrt{(2 - x)^{2} + 1}} d x & = - \int \frac{1}{\sqrt{t^{2} + 1}} d t \\ = - \log | t + \sqrt{t^{2} + 1} | + C \end{aligned} \begin{array}{l} [\int \frac{1}{\sqrt{x^{2} + a^{2}}} d t = \log | x + \sqrt{x^{2} + a^{2}} |] \end{array}= − log | 2 − x + √ (2 − x) 2 + 1 + C = log 1 (2 − x) + √ x 2 − 4 x + 5 + C}

\sqrt{Q4. Integrate the function :1 √ 9 − 25 x 2 1 \sqrt{9 - 25 x^{2}} \sqrt{}}

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\sqrt{Answer. Let 5x = t ∴ 5dx = dt⇒ ∫ 1 √ 9 − 25 x 2 d x = 15 ∫ 1 9 − t 2 d t = 15 ∫ 1 √ 32 − t 2 d t = 15 sin − 1 (t 3) + C = 15 sin − 1 (5 x 3) + C}

\sqrt{Q5. Integrate the function :3 x 1 + 2 x 4 3 x 1 + 2 x^{4}}

\sqrt{Answer.Let √ 2 x 2 = t ∴ 2 √ 2 x d x = d t ⇒ ∫ 3 x 1 + 2 x 4 d x = 3 2 √ 2 ∫ d t 1 + t 2 \begin{array}{l} Let \sqrt{2} x^{2} = t \\ ∴ 2 \sqrt{2} x d x = d t \\ \Rightarrow \int \frac{3 x}{1 + 2 x^{4}} d x = \frac{3}{2 \sqrt{2}} \int \frac{d t}{1 + t^{2}} \end{array}= 3 2 √ 2 [tan − 1 t] + C = 3 2 √ 2 tan − 1 (√ 2 x 2) + C}

\sqrt{Q6. Integrate the function : x 2 1 − x 6 \frac{x^{2}}{1 - x^{6}}}

\sqrt{Answer. Letx 3 = t x^{3} = t∴ 3 x 2 d x = d t ⇒ ∫ x 2 1 − x 6 d x = 13 ∫ d t 1 − t 2 = 13 [12 log ∣ ∣ ∣ 1 + t 1 − t ∣ ∣ ∣] + C \begin{aligned} ∴ 3 x^{2} d x = d t \\ \Rightarrow \int \frac{x^{2}}{1 - x^{6}} d x & = \frac{1}{3} \int \frac{d t}{1 - t^{2}} \\ = \frac{1}{3} [\frac{1}{2} \log | \frac{1 + t}{1 - t} |] + C \end{aligned}= 16 log ∣ ∣ 1 + x 3 1 − x 3 ∣ ∣ + C}

\sqrt{Q7. Integrate the function : x − 1 √ x 2 − 1}

\sqrt{Answer.∫ x − 1 √ x 2 − 1 d x = ∫ x √ x 2 − 1 d x − ∫ 1 √ x 2 − 1 d x . . . . (1) \int \frac{x - 1}{\sqrt{x^{2} - 1}} d x = \int \frac{x}{\sqrt{x^{2} - 1}} d x - \int \frac{1}{\sqrt{x^{2} - 1}} d x . . . . (1)For ∫ x √ x 2 − 1 d x, let x 2 − 1 = t ⇒ 2 x d x = d t ∴ ∫ x √ x 2 − 1 d x = 12 ∫ d t √ t \begin{array}{l} For \int \frac{x}{\sqrt{x^{2} - 1}} d x, let x^{2} - 1 = t \Rightarrow 2 x d x = d t \\ ∴ \int \frac{x}{\sqrt{x^{2} - 1}} d x = \frac{1}{2} \int \frac{d t}{\sqrt{t}} \end{array}= 12 ∫ t − 12 d t = 12 [2 t 12] = √ t = √ x 2 − 1 From ( 1), we obtain \begin{aligned} = \frac{1}{2} \int t^{- \frac{1}{2}} d t \\ = \frac{1}{2} [2 t^{\frac{1}{2}}] \\ = \sqrt{t} \\ = \sqrt{x^{2} - 1} \\ From ( 1), we obtain \end{aligned}∫ x − 1 √ x 2 − 1 d x = ∫ x √ x 2 − 1 d x − ∫ 1 √ x 2 − 1 d x [∫ 1 √ x 2 − a 2 d t = log ∣ ∣ x + √ x 2 − a 2 ∣ ∣] \int \frac{x - 1}{\sqrt{x^{2} - 1}} d x = \int \frac{x}{\sqrt{x^{2} - 1}} d x - \int \frac{1}{\sqrt{x^{2} - 1}} d x [\int \frac{1}{\sqrt{x^{2} - a^{2}}} d t = \log | x + \sqrt{x^{2} - a^{2}} |]= √ x 2 − 1 − log ∣ ∣ x + √ x 2 − 1 ∣ ∣ + C}

\sqrt{Q8. Integrate the function :x 2 √ x 6 + a 6 x^{2} \sqrt{x^{6} + a^{6}} \sqrt{}}

\sqrt{Answer.Let x 3 = t ⇒ 3 x 2 d x = d t ∴ ∫ x 2 √ x 6 + a 6 d x = 13 ∫ d t √ t 2 + (a 3) 2 \begin{array}{l} Let x^{3} = t \\ \Rightarrow 3 x^{2} d x = d t \\ ∴ \int \frac{x^{2}}{\sqrt{x^{6} + a^{6}}} d x = \frac{1}{3} \int \frac{d t}{\sqrt{t^{2} + {(a^{3})}^{2}}} \end{array}= 13 log ∣ ∣ t + √ t 2 + a 6 ∣ ∣ + C = 13 log ∣ ∣ x 3 + √ x 6 + a 6 ∣ ∣ + C}

\sqrt{Q9. Integrate the function :sec 2 x √ tan 2 x + 4 \sec^{2} x \sqrt{\tan^{2} x + 4} \sqrt{}}

\sqrt{Answer.Let tan x = t ∴ sec 2 x d x = d t \begin{array}{l} Let tan x = t \\ ∴ \sec^{2} x d x = d t \end{array}⇒ ∫ sec 2 x √ tan 2 x + 4 d x = ∫ d t √ t 2 + 22 = log ∣ ∣ t + √ t 2 + 4 ∣ ∣ + C = log ∣ ∣ tan x + √ tan 2 x + 4 ∣ ∣ + C}

\sqrt{Q10. Integrate the function :1 √ x 2 + 2 x + 2 1 \sqrt{x^{2} + 2 x + 2} \sqrt{}}

\sqrt{Answer.∫ 1 √ x 2 + 2 x + 2 d x = ∫ 1 √ (x + 1) 2 + (1) 2 d x Let x + 1 = t ∴ d x = d t ⇒ ∫ 1 √ x 2 + 2 x + 2 d x = ∫ 1 √ t 2 + 1 d t \begin{array}{l} \int \frac{1}{\sqrt{x^{2} + 2 x + 2}} d x = \int \frac{1}{\sqrt{(x + 1)^{2} + (1)^{2}}} d x \\ Let x + 1 = t \\ ∴ d x = d t \\ \Rightarrow \int \frac{1}{\sqrt{x^{2} + 2 x + 2}} d x = \int \frac{1}{\sqrt{t^{2} + 1}} d t \end{array}= log ∣ ∣ t + √ t 2 + 1 ∣ ∣ + C = log ∣ ∣ (x + 1) + √ (x + 1) 2 + 1 ∣ ∣ + C = log ∣ ∣ (x + 1) + √ x 2 + 2 x + 2 ∣ ∣ + C}

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\sqrt{Q11. Integrate the function :1 √ 9 x 2 + 6 x + 5 1 \sqrt{9 x^{2} + 6 x + 5} \sqrt{}}

\sqrt{Answer.∫ 1 9 x 2 + 6 x + 5 d x = ∫ 1 (3 x + 1) 2 + (2) 2 d x Let (3 x + 1) = t ∴ 3 d x = d t ⇒ ∫ 1 (3 x + 1) 2 + (2) 2 d x = 13 ∫ 1 t 2 + 22 d t \begin{array}{l} \int \frac{1}{9 x^{2} + 6 x + 5} d x = \int \frac{1}{(3 x + 1)^{2} + (2)^{2}} d x \\ Let (3 x + 1) = t \\ ∴ 3 d x = d t \\ \Rightarrow \int \frac{1}{(3 x + 1)^{2} + (2)^{2}} d x = \frac{1}{3} \int \frac{1}{t^{2} + 2^{2}} d t \end{array}= 13 [12 tan − 1 (t 2)] + C = 16 tan − 1 (3 x + 1 2) + C}

\sqrt{Q12. Integrate the function :1 √ 7 − 6 x − x 2 1 \sqrt{7 - 6 x - x^{2}} \sqrt{}}

\sqrt{Answer.7 − 6 x − x 2 can be written as 7 − (x 2 + 6 x + 9 − 9) Therefore, 7 − (x 2 + 6 x + 9 − 9) = 16 − (x 2 + 6 x + 9) = 16 − (x + 3) 2 \begin{array}{l} 7 - 6 x - x^{2} can be written as 7 - (x^{2} + 6 x + 9 - 9) \\ Therefore, \\ 7 - (x^{2} + 6 x + 9 - 9) \\ = 16 - (x^{2} + 6 x + 9) \\ = 16 - (x + 3)^{2} \end{array}= (4) 2 − (x + 3) 2 ∴ ∫ 1 √ 7 − 6 x − x 2 d x = ∫ 1 √ (4) 2 − (x + 3) 2 d x Let x + 3 = t ⇒ d x = d t ⇒ ∫ 1 √ (4) 2 − (x + 3) 2 d x = ∫ 1 √ (4) 2 − (t) 2 d t \begin{array}{l} = (4)^{2} - (x + 3)^{2} \\ ∴ \int \frac{1}{\sqrt{7 - 6 x - x^{2}}} d x = \int \frac{1}{\sqrt{(4)^{2} - (x + 3)^{2}}} d x \\ Let x + 3 = t \\ \Rightarrow d x = d t \\ \Rightarrow \int \frac{1}{\sqrt{(4)^{2} - (x + 3)^{2}}} d x = \int \frac{1}{\sqrt{(4)^{2} - (t)^{2}}} d t \end{array}= sin − 1 (t 4) + C = sin − 1 (x + 3 4) + C}

\sqrt{Q13. Integrate the function : 1 √ (x − 1) (x − 2)}

\sqrt{Answer.(x − 1) (x − 2) can be written as x 2 − 3 x + 2 Therefore, x 2 − 3 x + 2 = x 2 − 3 x + 94 − 94 + 2 = (x − 32) 2 − 14 = (x − 32) 2 − (12) 2 \begin{array}{l} (x - 1) (x - 2) can be written as x^{2} - 3 x + 2 \\ Therefore, \\ x^{2} - 3 x + 2 \\ = x^{2} - 3 x + \frac{9}{4} - \frac{9}{4} + 2 \\ = {(x - \frac{3}{2})}^{2} - \frac{1}{4} \\ = {(x - \frac{3}{2})}^{2} - {(\frac{1}{2})}^{2} \end{array}∴ ∫ 1 √ (x − 1) (x − 2) d x = ∫ 1 √ (x − 32) 2 − (12) 2 d x Let x − 32 = t ∴ d x = d t ⇒ ∫ 1 √ (x − 32) 2 − (12) 2 d x = ∫ 1 √ t 2 − (12) 2 d t \begin{array}{l} ∴ \int \frac{1}{\sqrt{(x - 1) (x - 2)}} d x = \int \frac{1}{\sqrt{{(x - \frac{3}{2})}^{2} - {(\frac{1}{2})}^{2}}} d x \\ Let x - \frac{3}{2} = t \\ ∴ d x = d t \\ \Rightarrow \int \frac{1}{\sqrt{{(x - \frac{3}{2})}^{2} - {(\frac{1}{2})}^{2}}} d x = \int \frac{1}{\sqrt{t^{2} - {(\frac{1}{2})}^{2}}} d t \end{array}= log ∣ ∣ ∣ t + √ t 2 − (12) 2 ∣ ∣ ∣ + C = log (x − 32) + √ x 2 − 3 x + 2 | + C}

\sqrt{Q14. Integrate the function :1 √ 8 + 3 x − x 2 1 \sqrt{8 + 3 x - x^{2}} \sqrt{}}

\sqrt{Answer.8 + 3 x − x 2 can be written as 8 − (x 2 − 3 x + 94 − 94) Therefore, 8 − (x 2 − 3 x + 94 − 94) = 414 − (x − 32) 2 \begin{array}{l} 8 + 3 x - x^{2} can be written as 8 - (x^{2} - 3 x + \frac{9}{4} - \frac{9}{4}) \\ Therefore, \\ 8 - (x^{2} - 3 x + \frac{9}{4} - \frac{9}{4}) \\ = \frac{41}{4} - {(x - \frac{3}{2})}^{2} \end{array}⇒ ∫ 1 √ 8 + 3 x − x 2 d x = ∫ 1 √ 414 − (x − 32) 2 d x Let x − 32 = t ∴ d x = d t ⇒ ∫ 1 √ 414 − (x − 32) 2 d x = ∫ 1 √ (√ 41 2) 2 − t 2 d t \begin{array}{l} \Rightarrow \int \frac{1}{\sqrt{8 + 3 x - x^{2}}} d x = \int \frac{1}{\sqrt{\frac{41}{4} - {(x - \frac{3}{2})}^{2}} d x} \\ Let x - \frac{3}{2} = t \\ ∴ d x = d t \\ \Rightarrow \int \frac{1}{\sqrt{\frac{41}{4} - {(x - \frac{3}{2})}^{2}}} d x = \int \frac{1}{\sqrt{{(\frac{\sqrt{41}}{2})}^{2} - t^{2}}} d t \end{array}= sin − 1 (t √ 41) + C = sin − 1 (x − 32 √ 41 2) + C = sin − 1 (2 x − 3 √ 41) + C}

\sqrt{Q15. Integrate the function : 1 √ (x − a) (x − b) \frac{1}{\sqrt{(x - a) (x - b)}} \sqrt{}}

\sqrt{Answer.(x − a) (x − b) can be written as x 2 − (a + b) x + a b Therefore, x 2 − (a + b) x + a b = x 2 − (a + b) x + (a + b) 2 4 − (a + b) 2 4 + a b = [x − (a + b 2)] 2 − (a − b) 2 4 \begin{array}{l} (x - a) (x - b) can be written as x^{2} - (a + b) x + a b \\ Therefore, \\ x^{2} - (a + b) x + a b \\ = x^{2} - (a + b) x + \frac{(a + b)^{2}}{4} - \frac{(a + b)^{2}}{4} + a b \\ = {[x - (\frac{a + b}{2})]}^{2} - \frac{(a - b)^{2}}{4} \end{array}⇒ ∫ 1 √ (x − a) (x − b) d x = ∫ 1 √ {x − (a + b 2)} 2 − (a − b 2) 2 d x Let x − (a + b 2) = t ∴ d x = d t ⇒ ∫ 1 √ {x − (a + b 2)} 2 − (a − b 2) 2 d x = ∫ 1 √ t 2 − (a − b 2) 2 d t \begin{array}{l} \Rightarrow \int \frac{1}{\sqrt{(x - a) (x - b)} d x} = \int \frac{1}{\sqrt{{x - (\frac{a + b}{2})}^{2} - {(\frac{a - b}{2})}^{2}}} d x \\ Let x - (\frac{a + b}{2}) = t \\ ∴ d x = d t \\ \Rightarrow \int \frac{1}{\sqrt{{x - (\frac{a + b}{2})}^{2} - {(\frac{a - b}{2})}^{2}} d x = \int \frac{1}{\sqrt{t^{2} - {(\frac{a - b}{2})}^{2}}} d t} \end{array}= log ∣ ∣ ∣ t + √ t 2 − (a − b 2) 2 ∣ ∣ ∣ + C = log | {x − (a + b 2)} + √ (x − a) (x − b) + C}

\sqrt{Q16. Integrate the function :4 x + 1 √ 2 x 2 + x − 3 4 x + 1 \sqrt{2 x^{2} + x - 3} \sqrt{}}

\sqrt{Answer.Let 4 x + 1 = A d d x (2 x 2 + x − 3) + B ⇒ 4 x + 1 = A (4 x + 1) + B ⇒ 4 x + 1 = 4 A x + A + B \begin{array}{l} Let 4 x + 1 = A \frac{d}{d x} (2 x^{2} + x - 3) + B \\ \Rightarrow 4 x + 1 = A (4 x + 1) + B \\ \Rightarrow 4 x + 1 = 4 A x + A + B \end{array}Equating the coefficients of x and constant term on both sides, we obtain 4A = 4 ⇒ A = 1 A + B = 1 ⇒ B = 0Let 2 x 2 + x − 3 = t ∴ (4 x + 1) d x = d t \begin{array}{l} Let 2 x^{2} + x - 3 = t \\ ∴ (4 x + 1) d x = d t \end{array}⇒ ∫ 4 x + 1 √ 2 x 2 + x − 3 d x = ∫ 1 √ t d t = 2 √ t + C = 2 √ 2 x 2 + x − 3 + C}

\sqrt{Q17. Integrate the function :x + 2 √ x 2 − 1 x + 2 \sqrt{x^{2} - 1} \sqrt{}}

\sqrt{Answer.Let x + 2 = A d d x (x 2 − 1) + B … (1) ⇒ x + 2 = A (2 x) + B \begin{array}{ll} Let x + 2 = A \frac{d}{d x} (x^{2} - 1) + B & \dots (1) \\ \Rightarrow x + 2 = A (2 x) + B \end{array}Equating the coefficients of x and constant term on both sides, we obtain2 A = 1 ⇒ A = 12 B = 2 From (1), we obtain \begin{array}{l} 2 A = 1 \Rightarrow A = \frac{1}{2} \\ B = 2 \\ From (1), we obtain \end{array}(x + 2) = 12 (2 x) + 2 Then, ∫ x + 2 √ x 2 − 1 d x = ∫ 12 (2 x) + 2 √ x 2 − 1 d x = 12 ∫ 2 x √ x 2 − 1 d x + ∫ 2 √ x 2 − 1 d x . . . (2) \begin{aligned} (x + 2) = \frac{1}{2} (2 x) + 2 \\ Then, \int \frac{x + 2}{\sqrt{x^{2} - 1}} d x & = \int \frac{\frac{1}{2} (2 x) + 2}{\sqrt{x^{2} - 1}} d x \\ = \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2} - 1}} d x + \int \frac{2}{\sqrt{x^{2} - 1}} d x . . . (2) \end{aligned}In 12 ∫ 2 x √ x 2 − 1 d x, let x 2 − 1 = t ⇒ 2 x d x = d t 12 ∫ 2 x √ x 2 − 1 d x = 12 ∫ d t √ t = 12 [2 √ t] = √ t = √ x 2 − 1 \begin{aligned} In \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2} - 1}} d x, & let x^{2} - 1 = t \Rightarrow 2 x d x = d t \\ \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2} - 1}} d x & = \frac{1}{2} \int \frac{d t}{\sqrt{t}} \\ = \frac{1}{2} [2 \sqrt{t}] \\ = \sqrt{t} \\ = \sqrt{x^{2} - 1} \end{aligned}Then, ∫ 2 √ x 2 − 1 d x = 2 ∫ 1 √ x 2 − 1 d x = 2 log ∣ ∣ x + √ x 2 − 1 ∣ ∣ From equation (2), we obtain ∫ x + 2 √ x 2 − 1 d x = √ x 2 − 1 + 2 log ∣ ∣ x + √ x 2 − 1 ∣ ∣ + C}

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\sqrt{Q18. Integrate the function :5 x − 2 1 + 2 x + 3 x 2 5 x - 2 1 + 2 x + 3 x^{2}}

\sqrt{Answer.Let 5 x − 2 = A d d x (1 + 2 x + 3 x 2) + B ⇒ 5 x − 2 = A (2 + 6 x) + B \begin{array}{l} Let 5 x - 2 = A \frac{d}{d x} (1 + 2 x + 3 x^{2}) + B \\ \Rightarrow 5 x - 2 = A (2 + 6 x) + B \end{array}Equating the coefficient of x and constant term on both sides, we obtain5 = 6 A ⇒ A = 56 2 A + B = − 2 ⇒ B = − 113 ∴ 5 x − 2 = 56 (2 + 6 x) + (− 113) ⇒ ∫ 5 x − 2 1 + 2 x + 3 x 2 d x = ∫ 56 (2 + 6 x − 113) 1 + 2 x + 3 x 2 d x \begin{array}{l} 5 = 6 A \Rightarrow A = \frac{5}{6} \\ 2 A + B = - 2 \Rightarrow B = - \frac{11}{3} \\ ∴ 5 x - 2 = \frac{5}{6} (2 + 6 x) + (- \frac{11}{3}) \\ \Rightarrow \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x = \int \frac{56 (2 + 6 x - \frac{11}{3})}{1 + 2 x + 3 x^{2}} d x \end{array}= 56 ∫ 2 + 6 x 1 + 2 x + 3 x 2 d x − 113 ∫ 1 1 + 2 x + 3 x 2 d x Let I 1 = ∫ 2 + 6 x 1 + 2 x + 3 x 2 d x and I 2 = ∫ 1 1 + 2 x + 3 x 2 d x ∴ ∫ 5 x − 2 1 + 2 x + 3 x 2 d x = 56 I 1 − 113 I 2 . . . . . . (1) \begin{array}{l} = \frac{5}{6} \int \frac{2 + 6 x}{1 + 2 x + 3 x^{2}} d x - \frac{11}{3} \int \frac{1}{1 + 2 x + 3 x^{2}} d x \\ Let I_{1} = \int \frac{2 + 6 x}{1 + 2 x + 3 x^{2}} d x and I_{2} = \int \frac{1}{1 + 2 x + 3 x^{2}} d x \\ ∴ \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x = \frac{5}{6} I_{1} - \frac{11}{3} I_{2} . . . . . . (1) \end{array}I 1 = ∫ 2 + 6 x 1 + 2 x + 3 x 2 d x Let 1 + 2 x + 3 x 2 ⇒ (2 + 6 x) d x = d t ∴ I 1 = ∫ d t t I 1 = log | t | I 1 = log ∣ ∣ 1 + 2 x + 3 x 2 ∣ ∣ \begin{array}{l} I_{1} = \int \frac{2 + 6 x}{1 + 2 x + 3 x^{2}} d x \\ Let 1 + 2 x + 3 x^{2} \\ \Rightarrow (2 + 6 x) d x = d t \\ ∴ I_{1} = \int \frac{d t}{t} \\ I_{1} = \log | t | \\ I_{1} = \log | 1 + 2 x + 3 x^{2} | \end{array}I 2 = ∫ 1 1 + 2 x + 3 x 2 d x I_{2} = \int \frac{1}{1 + 2 x + 3 x^{2}} d x1 + 2 x + 3 x 2 can be written as 1 + 3 (x 2 + 23 x) Therefore, 1 + 3 (x 2 + 23 x) = 1 + 3 (x 2 + 23 x + 19 − 19) \begin{array}{l} 1 + 2 x + 3 x^{2} can be written as 1 + 3 (x^{2} + \frac{2}{3} x) \\ Therefore, \\ 1 + 3 (x^{2} + \frac{2}{3} x) \\ = 1 + 3 (x^{2} + \frac{2}{3} x + \frac{1}{9} - \frac{1}{9}) \end{array}= 1 + 3 (x + 13) 2 − 13 = 23 + 3 (x + 13) 2 = 3 [(x + 13) 2 + 29] = 3 [(x + 13) 2 + (√ 2 3) 2] \begin{array}{l} = 1 + 3 {(x + \frac{1}{3})}^{2} - \frac{1}{3} \\ = \frac{2}{3} + 3 {(x + \frac{1}{3})}^{2} \\ = 3 [{(x + \frac{1}{3})}^{2} + \frac{2}{9}] \\ = 3 [{(x + \frac{1}{3})}^{2} + {(\frac{\sqrt{2}}{3})}^{2}] \end{array}I 2 = 13 ∫ 1 [(x + 13) 2 + (√ 2 3) 2] d x = 13 ⎡ ⎢ ⎣ 1 √ 2 3 tan − 1 ⎛ ⎜ ⎝ x + 13 √ 2 3 ⎞ ⎟ ⎠ ⎤ ⎥ ⎦ \begin{aligned} I_{2} & = \frac{1}{3} \int \frac{1}{{[{(x + \frac{1}{3})}^{2} + {(\frac{\sqrt{2}}{3})}^{2}]}^{d x}} \\ = \frac{1}{3} [\frac{1}{\frac{\sqrt{2}}{3}} \tan^{- 1} (\frac{x + \frac{1}{3}}{\frac{\sqrt{2}}{3}})] \end{aligned}= 13 [3 √ 2 tan − 1 (3 x + 1 √ 2)] = 1 √ 2 tan − 1 (3 x + 1 √ 2 . . (3)) \begin{aligned} = \frac{1}{3} [\frac{3}{\sqrt{2}} \tan^{- 1} (\frac{3 x + 1}{\sqrt{2}})] \\ = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{3 x + 1}{\sqrt{2} . . (3)}) \end{aligned}Substituting equations (2) and (3) in equation (1), we obtain∫ 5 x − 2 1 + 2 x + 3 x 2 d x = 56 [log | 1 + 2 x + 3 x 2 |] − 113 [1 √ 2 tan − 1 (3 x + 1 √ 2)] + C = 56 log ∣ ∣ 1 + 2 x + 3 x 2 ∣ ∣ − 11 3 √ 2 tan − 1 (3 x + 1 √ 2) + C}

\sqrt{Q19. Integrate the function :6 x + 7 √ (x − 5) (x − 4) 6 x + 7 \sqrt{(x - 5) (x - 4)} \sqrt{}}

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\sqrt{Answer.6 x + 7 √ (x − 5) (x − 4) = 6 x + 7 √ x 2 − 9 x + 20 Let 6 x + 7 = A d d x (x 2 − 9 x + 20) + B ⇒ 6 x + 7 = A (2 x − 9) + B \begin{array}{l} \frac{6 x + 7}{\sqrt{(x - 5) (x - 4)}} = \frac{6 x + 7}{\sqrt{x^{2} - 9 x + 20}} \\ Let 6 x + 7 = A \frac{d}{d x} (x^{2} - 9 x + 20) + B \\ \Rightarrow 6 x + 7 = A (2 x - 9) + B \end{array}Equating the coefficients of x and constant term, we obtain 2A = 6 ⇒ A = 3 −9A + B = 7 ⇒ B = 34 ∴ 6x + 7 = 3 (2x − 9) + 34∫ 6 x + 7 √ x 2 − 9 x + 20 = ∫ 3 (2 x − 9) + 34 √ x 2 − 9 x + 20 d x = 3 ∫ 2 x − 9 √ x 2 − 9 x + 20 d x + 34 ∫ 1 √ x 2 − 9 x + 20 d x \begin{aligned} \int \frac{6 x + 7}{\sqrt{x^{2} - 9 x + 20}} & = \int \frac{3 (2 x - 9) + 34}{\sqrt{x^{2} - 9 x + 20}} d x \\ = 3 \int \frac{2 x - 9}{\sqrt{x^{2} - 9 x + 20}} d x + 34 \int \frac{1}{\sqrt{x^{2} - 9 x + 20}} d x \end{aligned}Let I 1 = ∫ 2 x − 9 √ x 2 − 9 x + 20 d x and I 2 = ∫ 1 √ x 2 − 9 x + 20 d x ∴ ∫ 6 x + 7 √ x 2 − 9 x + 20 = 3 I 1 + 34 I 2 . . . . (1) \begin{array}{l} Let I_{1} = \int \frac{2 x - 9}{\sqrt{x^{2} - 9 x + 20}} d x and I_{2} = \int \frac{1}{\sqrt{x^{2} - 9 x + 20}} d x \\ ∴ \int \frac{6 x + 7}{\sqrt{x^{2} - 9 x + 20}} = 3 I_{1} + 34 I_{2} . . . . (1) \end{array}Then, I 1 = ∫ 2 x − 9 √ x 2 − 9 x + 20 d x Let x 2 − 9 x + 20 = t ⇒ (2 x − 9) d x = d t ⇒ I 1 = d t √ t \begin{array}{l} Then, \\ I_{1} = \int \frac{2 x - 9}{\sqrt{x^{2} - 9 x + 20}} d x \\ Let x^{2} - 9 x + 20 = t \\ \Rightarrow (2 x - 9) d x = d t \\ \Rightarrow I_{1} = \frac{d t}{\sqrt{t}} \end{array}I 1 = 2 √ t I 1 = 2 √ x 2 − 9 x + 20 and I 2 = ∫ 1 √ x 2 − 9 x + 20 d x \begin{array}{l} I_{1} = 2 \sqrt{t} \\ I_{1} = 2 \sqrt{x^{2} - 9 x + 20} \\ and I_{2} = \int \frac{1}{\sqrt{x^{2} - 9 x + 20}} d x \end{array}x 2 − 9 x + 20 can be written as x 2 − 9 x + 20 + 814 − 814 Therefore, x 2 − 9 x + 20 + 814 − 814 = (x − 92) 2 − 14 \begin{array}{l} x^{2} - 9 x + 20 can be written as x^{2} - 9 x + 20 + \frac{81}{4} - \frac{81}{4} \\ Therefore, \\ x^{2} - 9 x + 20 + \frac{81}{4} - \frac{81}{4} \\ = {(x - \frac{9}{2})}^{2} - \frac{1}{4} \end{array}= (x − 92) 2 − (12) 2 ⇒ I 2 = ∫ √ (x − 92) 2 − (12) 2 d x \begin{aligned} = {(x - \frac{9}{2})}^{2} - {(\frac{1}{2})}^{2} \\ \Rightarrow I_{2} = \int & \sqrt{{(x - \frac{9}{2})}^{2} - {(\frac{1}{2})}^{2}} d x \end{aligned}I 2 = log (x − 92) + √ x 2 − 9 x + 20 | . . . . . . . . (3) Substituting equations (2) and (3) in (1), we obtain ∫ 6 x + 7 √ x 2 − 9 x + 20 d x = 3 [2 √ x 2 − 9 x + 20] + 34 log [(x − 92) + √ x 2 − 9 x + 20] + C \begin{array}{l} I_{2} = \log (x - \frac{9}{2}) + \sqrt{x^{2} - 9 x + 20} | . . . . . . . . (3) \\ Substituting equations (2) and (3) in (1), we obtain \\ \int \frac{6 x + 7}{\sqrt{x^{2} - 9 x + 20} d x = 3 [2 \sqrt{x^{2} - 9 x + 20}] + 34 \log [(x - \frac{9}{2}) + \sqrt{x^{2} - 9 x + 20}] + C} \end{array}= 6 √ x 2 − 9 x + 20 + 34 log [(x − 92) + √ x 2 − 9 x + 20] + C}

\sqrt{Q20. Integrate the function :x + 2 √ 4 x − x 2 x + 2 \sqrt{4 x - x^{2}} \sqrt{}}

\sqrt{}

\sqrt{Answer.Let x + 2 = A d d x (4 x − x 2) + B ⇒ x + 2 = A (4 − 2 x) + B \begin{array}{l} Let x + 2 = A \frac{d}{d x} (4 x - x^{2}) + B \\ \Rightarrow x + 2 = A (4 - 2 x) + B \end{array}Equating the coefficients of x and constant term on both sides, we obtain− 2 A = 1 ⇒ A = − 12 4 A + B = 2 ⇒ B = 4 ⇒ (x + 2) = − 12 (4 − 2 x) + 4 ∴ ∫ x + 2 √ 4 x − x 2 d x = ∫ − 12 (4 − 2 x) + 4 √ 4 x − x 2 d x \begin{array}{l} - 2 A = 1 \Rightarrow A = - \frac{1}{2} \\ 4 A + B = 2 \Rightarrow B = 4 \\ \Rightarrow (x + 2) = - \frac{1}{2} (4 - 2 x) + 4 \\ ∴ \int \frac{x + 2}{\sqrt{4 x - x^{2}}} d x = \int \frac{- \frac{1}{2} (4 - 2 x) + 4}{\sqrt{4 x - x^{2}}} d x \end{array}= − 12 ∫ 4 − 2 x √ 4 x − x 2 d x + 4 ∫ 1 √ 4 x − x 2 d x Let I 1 = ∫ 4 − 2 x √ 4 x − x 2 d x and I 2 ∫ 1 √ 4 x − x 2 d x ∴ ∫ x + 2 √ 4 x − x 2 d x = − 12 I 1 + 4 I 2 \begin{array}{l} = - \frac{1}{2} \int \frac{4 - 2 x}{\sqrt{4 x - x^{2}}} d x + 4 \int \frac{1}{\sqrt{4 x - x^{2}}} d x \\ Let I_{1} = \int \frac{4 - 2 x}{\sqrt{4 x - x^{2}}} d x and I_{2} \int \frac{1}{\sqrt{4 x - x^{2}}} d x \\ ∴ \int \frac{x + 2}{\sqrt{4 x - x^{2}}} d x = - \frac{1}{2} I_{1} + 4 I_{2} \end{array}Then, I 1 = ∫ 4 − 2 x √ 4 x − x 2 d x Let 4 x − x 2 = t ⇒ (4 − 2 x) d x = d t ⇒ I 1 = ∫ d t √ t = 2 √ t = 2 √ 4 x − x 2 \begin{array}{l} Then, I_{1} = \int \frac{4 - 2 x}{\sqrt{4 x - x^{2}}} d x \\ Let 4 x - x^{2} = t \\ \Rightarrow (4 - 2 x) d x = d t \\ \Rightarrow I_{1} = \int_{\sqrt{t}}^{d t} = 2 \sqrt{t} = 2 \sqrt{4 x - x^{2}} \end{array}I 2 = ∫ 1 √ 4 x − x 2 d x ⇒ 4 x − x 2 = − (− 4 x + x 2) = (− 4 x + x 2 + 4 − 4) = 4 − (x − 2) 2 ∴ I 2 = ∫ 1 √ (2) 2 − (x − 2) 2 d x = sin − 1 (x − 2 2) \begin{aligned} I_{2} = \int \frac{1}{\sqrt{4 x - x^{2}}} d x \\ \Rightarrow 4 x - x^{2} & = - (- 4 x + x^{2}) \\ = (- 4 x + x^{2} + 4 - 4) \\ = 4 - (x - 2)^{2} \\ ∴ I_{2} = \int \frac{1}{\sqrt{(2)^{2} - (x - 2)^{2}}} d x & = \sin^{- 1} (\frac{x - 2}{2}) \end{aligned}Using equations (2) and (3) in (1), we obtain∫ x + 2 √ 4 x − x 2 d x = − 12 (2 √ 4 x − x 2) + 4 sin − 1 (x − 2 2) + C = − √ 4 x − x 2 + 4 sin − 1 (x − 2 2) + C}

\sqrt{Q21. Integrate the function :x + 2 √ x 2 + 2 x + 3 x + 2 \sqrt{x^{2} + 2 x + 3} \sqrt{}}

\sqrt{Answer.∫ (x + 2) √ x 2 + 2 x + 3 d x = 12 ∫ 2 (x + 2) √ x 2 + 2 x + 3 d x = 12 ∫ 2 x + 4 √ x 2 + 2 x + 3 d x = 12 ∫ 2 x + 2 √ x 2 + 2 x + 3 d x + 12 ∫ 2 √ x 2 + 2 x + 3 d x \begin{aligned} \int \frac{(x + 2)}{\sqrt{x^{2} + 2 x + 3}} d x & = \frac{1}{2} \int \frac{2 (x + 2)}{\sqrt{x^{2} + 2 x + 3}} d x \\ = \frac{1}{2} \int \frac{2 x + 4}{\sqrt{x^{2} + 2 x + 3}} d x \\ = \frac{1}{2} \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x + 3}} d x + \frac{1}{2} \int \frac{2}{\sqrt{x^{2} + 2 x + 3}} d x \end{aligned}= 12 ∫ 2 x + 2 √ x 2 + 2 x + 3 d x + ∫ 1 √ x 2 + 2 x + 3 d x Let I 1 = ∫ 2 x + 2 √ x 2 + 2 x + 3 d x and I 2 = ∫ 1 √ x 2 + 2 x + 3 d x ∴ ∫ x + 2 √ x 2 + 2 x + 3 d x = 12 I 1 + I 2 . . . . . (1) \begin{array}{l} = \frac{1}{2} \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x + 3}} d x + \int \frac{1}{\sqrt{x^{2} + 2 x + 3}} d x \\ Let I_{1} = \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x + 3}} d x and I_{2} = \int \frac{1}{\sqrt{x^{2} + 2 x + 3}} d x \\ ∴ \int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} d x = \frac{1}{2} I_{1} + I_{2} . . . . . (1) \end{array}Then, I 1 = ∫ 2 x + 2 √ x 2 + 2 x + 3 d x Let x 2 + 2 x + 3 = t ⇒ (2 x + 2) d x = d t \begin{array}{l} Then, I_{1} = \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x + 3}} d x \\ Let x^{2} + 2 x + 3 = t \\ \Rightarrow (2 x + 2) d x = d t \end{array}I 1 = ∫ d t √ t = 2 √ t = 2 √ x 2 + 2 x + 3 . . . (2) I 2 = ∫ 1 √ x 2 + 2 x + 3 d x \begin{array}{l} I_{1} = \int \frac{d t}{\sqrt{t}} = 2 \sqrt{t} = 2 \sqrt{x^{2} + 2 x + 3} . . . (2) \\ I_{2} = \int \frac{1}{\sqrt{x^{2} + 2 x + 3}} d x \end{array}⇒ x 2 + 2 x + 3 = x 2 + 2 x + 1 + 2 = (x + 1) 2 + (√ 2) 2 ∴ I 2 = ∫ 1 √ (x + 1) 2 + (√ 2) 2 d x = log (x + 1) + √ x 2 + 2 x + 3 | . . . (3) \begin{array}{l} \Rightarrow x^{2} + 2 x + 3 = x^{2} + 2 x + 1 + 2 = (x + 1)^{2} + (\sqrt{2})^{2} \\ ∴ I_{2} = \int \frac{1}{\sqrt{(x + 1)^{2} + (\sqrt{2})^{2}}} d x = \log (x + 1) + \sqrt{x^{2} + 2 x + 3} | . . . (3) \end{array}Using equations (2) and (3) in (1), we obtain ∫ x + 2 √ x 2 + 2 x + 3 d x = 12 [2 √ x 2 + 2 x + 3] + log (x + 1) + √ x 2 + 2 x + 3 | + C = √ x 2 + 2 x + 3 + log (x + 1) + √ x 2 + 2 x + 3 | + C}

Q22. Integrate the function : $\frac{x + 3}{x^{2} - 2 x - 5}$

\sqrt{Answer.Let (x + 3) = A d d x (x 2 − 2 x − 5) + B (x + 3) = A (2 x − 2) + B Equating the coefficients of x and constant term on both sides, we obtain \begin{array}{l} Let (x + 3) = A \frac{d}{d x} (x^{2} - 2 x - 5) + B \\ (x + 3) = A (2 x - 2) + B \\ Equating the coefficients of x and constant term on both sides, we obtain \end{array}2 A = 1 ⇒ A = 12 − 2 A + B = 3 ⇒ B = 4 ∴ (x + 3) = 12 (2 x − 2) + 4 \begin{array}{l} 2 A = 1 \Rightarrow A = \frac{1}{2} \\ - 2 A + B = 3 \Rightarrow B = 4 \\ ∴ (x + 3) = \frac{1}{2} (2 x - 2) + 4 \end{array}⇒ ∫ x + 3 x 2 − 2 x − 5 d x = ∫ 12 (2 x − 2) + 4 x 2 − 2 x − 5 d x = 12 ∫ 2 x − 2 x 2 − 2 x − 5 d x + 4 ∫ 1 x 2 − 2 x − 5 d x \begin{aligned} \Rightarrow \int \frac{x + 3}{x^{2} - 2 x - 5} d x & = \int \frac{\frac{1}{2} (2 x - 2) + 4}{x^{2} - 2 x - 5} d x \\ = \frac{1}{2} \int \frac{2 x - 2}{x^{2} - 2 x - 5} d x + 4 \int \frac{1}{x^{2} - 2 x - 5} d x \end{aligned}Let I 1 = ∫ 2 x − 2 x 2 − 2 x − 5 d x and I 2 = ∫ 1 x 2 − 2 x − 5 d x ∴ ∫ x + 3 (x 2 − 2 x − 5) d x = 12 I 1 + 4 I 2 Then, I 1 = ∫ 2 x − 2 x 2 − 2 x − 5 d x \begin{array}{l} Let I_{1} = \int \frac{2 x - 2}{x^{2} - 2 x - 5} d x and I_{2} = \int \frac{1}{x^{2} - 2 x - 5} d x \\ ∴ \int \frac{x + 3}{(x^{2} - 2 x - 5)} d x = \frac{1}{2} I_{1} + 4 I_{2} \\ Then, I_{1} = \int \frac{2 x - 2}{x^{2} - 2 x - 5} d x \end{array}Let x 2 − 2 x − 5 = t ⇒ (2 x − 2) d x = d t ⇒ I 1 = ∫ d t t = log | t | = log ∣ ∣ x 2 − 2 x − 5 ∣ ∣ \begin{array}{l} Let x^{2} - 2 x - 5 = t \\ \Rightarrow (2 x - 2) d x = d t \\ \Rightarrow I_{1} = \int \frac{d t}{t} = \log | t | = \log | x^{2} - 2 x - 5 | \end{array}I 2 = ∫ x 2 − 2 x − 5 d x = ∫ 1 (x 2 − 2 x + 1) − 6 d x = ∫ (x − 1) 2 + (√ 6) 2 d x = 1 2 √ 6 log (x − 1 − √ 6 x − 1 + √ 6) \begin{aligned} I_{2} & = \int_{x^{2} - 2 x - 5} d x \\ = \int \frac{1}{(x^{2} - 2 x + 1) - 6} d x \\ = \int \frac{1}{(x - 1)^{2} + (\sqrt{6})^{2}} d x \\ = \frac{1}{2 \sqrt{6}} \log (\frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}}) \end{aligned}Substituting (2) and (3) in (1), we obtain ∫ x + 3 x 2 − 2 x − 5 d x = 12 log ∣ ∣ x 2 − 2 x − 5 ∣ ∣ + 4 2 √ 6 log ∣ ∣ x − 1 − √ 6 x − 1 + √ 6 ∣ ∣ + C \begin{array}{l} Substituting (2) and (3) in (1), we obtain \\ \int \frac{x + 3}{x^{2} - 2 x - 5} d x = \frac{1}{2} \log | x^{2} - 2 x - 5 | + \frac{4}{2 \sqrt{6}} \log | \frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}} | + C \end{array}= 12 log ∣ ∣ x 2 − 2 x − 5 ∣ ∣ + 2 √ 6 log ∣ ∣ x − 1 − √ 6 x − 1 + √ 6 ∣ ∣ + C}

\sqrt{}

\sqrt{Q23. Integrate the function :5 x + 3 √ x 2 + 4 x + 10 5 x + 3 \sqrt{x^{2} + 4 x + 10} \sqrt{}}

\sqrt{}

\sqrt{Answer.Let 5 x + 3 = A d d x (x 2 + 4 x + 10) + B ⇒ 5 x + 3 = A (2 x + 4) + B Equating the coefficients of x and constant term, we obtain 2 A = 5 ⇒ A = 52 \begin{array}{l} Let 5 x + 3 = A \frac{d}{d x} (x^{2} + 4 x + 10) + B \\ \Rightarrow 5 x + 3 = A (2 x + 4) + B \\ Equating the coefficients of x and constant term, we obtain \\ 2 A = 5 \Rightarrow A = \frac{5}{2} \end{array}4 A + B = 3 ⇒ B = − 7 ∴ 5 x + 3 = 52 (2 x + 4) − 7 ⇒ ∫ 5 x + 3 √ x 2 + 4 x + 10 d x = ∫ 52 (2 x + 4) − 7 √ x 2 + 4 x + 10 d x \begin{array}{l} 4 A + B = 3 \Rightarrow B = - 7 \\ ∴ 5 x + 3 = \frac{5}{2} (2 x + 4) - 7 \\ \Rightarrow \int \frac{5 x + 3}{\sqrt{x^{2} + 4 x + 10}} d x = \int \frac{\frac{5}{2} (2 x + 4) - 7}{\sqrt{x^{2} + 4 x + 10}} d x \end{array}= 52 ∫ 2 x + 4 √ x 2 + 4 x + 10 d x − 7 ∫ 1 √ x 2 + 4 x + 10 d x Let I 1 = ∫ 2 x + 4 √ x 2 + 4 x + 10 d x and I 2 = ∫ 1 √ x 2 + 4 x + 10 d x \begin{matrix} = \frac{5}{2} \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 10}} d x - 7 \int \frac{1}{\sqrt{x^{2} + 4 x + 10}} d x \\ Let I_{1} = \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 10}} d x and I_{2} = \int \frac{1}{\sqrt{x^{2} + 4 x + 10}} d x \end{matrix}∴ ∫ 5 x + 3 √ x 2 + 4 x + 10 d x = 52 I 1 − 7 I 2 Then I 1 = ∫ 2 x + 4 √ x 2 + 4 x + 10 d x Let x 2 + 4 x + 10 = t ∴ (2 x + 4) d x = d t \begin{array}{l} ∴ \int \frac{5 x + 3}{\sqrt{x^{2} + 4 x + 10}} d x = \frac{5}{2} I_{1} - 7 I_{2} \\ Then I_{1} = \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 10}} d x \\ Let x^{2} + 4 x + 10 = t \\ ∴ (2 x + 4) d x = d t \end{array}⇒ I 1 = ∫ d t t = 2 √ t = 2 √ x 2 + 4 x + 10 I 2 = ∫ 1 √ x 2 + 4 x + 10 d x = ∫ 1 √ (x 2 + 4 x + 4) + 6 d x \begin{array}{l} \Rightarrow I_{1} = \int \frac{d t}{t} = 2 \sqrt{t} = 2 \sqrt{x^{2} + 4 x + 10} \\ I_{2} = \int \frac{1}{\sqrt{x^{2} + 4 x + 10}} d x \\ = \int \frac{1}{\sqrt{(x^{2} + 4 x + 4) + 6}} d x \end{array}= ∫ 1 (x + 2) 2 + (√ 6) 2 d x = log ∣ ∣ (x + 2) √ x 2 + 4 x + 10 ∣ ∣ \begin{aligned} = \int \frac{1}{(x + 2)^{2} + (\sqrt{6})^{2}} d x \\ = \log | (x + 2) \sqrt{x^{2} + 4 x + 10} | \end{aligned}Using equations (2) and (3) in (1), we obtain∫ 5 x + 3 √ x 2 + 4 x + 10 d x = 52 [2 √ x 2 + 4 x + 10] − 7 log (x + 2) + √ x 2 + 4 x + 10 + C = 5 √ x 2 + 4 x + 10 − 7 log (x + 2) + √ x 2 + 4 x + 10 | + C}

\sqrt{Q24. Choose the correct answer ∫ d x x 2 + 2 x + 2 \int \frac{d x}{x^{2} + 2 x + 2} equals (A) x tan − 1 (x + 1) + C (B) tan − 1 (x + 1) + C (C) (x + 1) tan − 1 x + C (D) tan − 1 x + C}

\sqrt{}

\sqrt{Answer. ∫ d x x 2 + 2 x + 2 = ∫ d x (x 2 + 2 x + 1) + 1 = ∫ 1 (x + 1) 2 + (1) 2 d x = [tan − 1 (x + 1)] + C \begin{aligned} \int \frac{d x}{x^{2} + 2 x + 2} & = \int \frac{d x}{(x^{2} + 2 x + 1) + 1} \\ = \int \frac{1}{(x + 1)^{2} + (1)^{2}} d x \\ = [\tan^{- 1} (x + 1)] + C \end{aligned}Hence, the correct Answer is B.}

\sqrt{}

\sqrt{Q25. Choose the correct answer ∫ d x √ 9 x − 4 x 2 \int \frac{d x}{\sqrt{9 x - 4 x^{2}}} equals (A) 19 sin − 1 (9 x − 8 8) + C (B) 12 sin − 1 (8 x − 9 9) + C (A) \frac{1}{9} \sin^{- 1} (\frac{9 x - 8}{8}) + C (B) \frac{1}{2} \sin^{- 1} (\frac{8 x - 9}{9}) + C(C) 13 sin − 1 (9 x − 8 8) + C (D) 12 sin − 1 (9 x − 8 9) + C (C) \frac{1}{3} \sin^{- 1} (\frac{9 x - 8}{8}) + C (D) \frac{1}{2} \sin^{- 1} (\frac{9 x - 8}{9}) + C Answer. ∫ d x √ 9 x − 4 x 2 = ∫ 1 √ − 4 (x 2 − 94 x) d x d x = ∫ 1 − 4 (x 2 − 94 x + 8164 − 8164) d x \begin{aligned} \int \frac{d x}{\sqrt{9 x - 4 x^{2}}} \\ = \int \frac{1}{\sqrt{- 4 (x^{2} - \frac{9}{4} x)} d x} d x \\ = \int \frac{1}{- 4 {(x^{2} - \frac{9}{4} x + \frac{81}{64} - \frac{81}{64})}^{d x}} \end{aligned} = ∫ 1 √ − 4 [(x − 98) 2 − (98) 2] d x = 12 ∫ 1 √ (98) 2 − (x − 98) 2 d x \begin{aligned} = \int \frac{1}{\sqrt{- 4 [{(x - \frac{9}{8})}^{2} - {(\frac{9}{8})}^{2}]} d x} \\ = \frac{1}{2} \int \frac{1}{\sqrt{{(\frac{9}{8})}^{2} - {(x - \frac{9}{8})}^{2}}} d x \end{aligned} = 12 [sin − 1 (x − 98 98)] + C (∫ d y √ a 2 − y 2 = sin − 1 y a + C) = \frac{1}{2} [\sin^{- 1} (\frac{x - \frac{9}{8}}{\frac{9}{8}})] + C (\int \frac{d y}{\sqrt{a^{2} - y^{2}}} = \sin^{- 1} \frac{y}{a} + C) = 12 sin − 1 (8 x − 9 9) + C Hence, the correct Answer is B.}

\sqrt{}

NCERT Solutions Class 12 maths Chapter-7 (Integrals)Exercise 7.4

NCERT Solutions Class 12 maths Chapter-7 (Integrals)Exercise 7.4

Exercise 7.4

Chapter-7 (integrals)

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NCERT Solutions Class 12 maths Chapter-7 (Integrals)Exercise 7.4

NCERT Solutions Class 12 maths Chapter-7 (Integrals)Exercise 7.4

Exercise 7.4

Chapter-7 (integrals)

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