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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

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

Exercise 7.4

Q1. Integrate the function : 


Q2. Integrate the function : 

Answer. Let 2x = t ∴ 2dx = dt 11+4x2dx=12dt1+t2=12[log|t+t2+1|]+C[1x2+a2dt=log|x+x2+a2|] 

Q3. Integrate the function : 1(2x)2+1

Answer. Let 2 − x = t ⇒ −dx = dt 1(2x)2+1dx=1t2+1dt=log|t+t2+1|+C[1x2+a2dt=log|x+x2+a2|] 
Q4. Integrate the function : 1925x2

Answer.Let 5x = t ∴ 5dx = dt 
Q5. Integrate the function : 3x1+2x4

Answer.  Let 2x2=t22xdx=dt3x1+2x4dx=322dt1+t2 
Q6. Integrate the function : x21x6

Answer. Let x3=t 3x2dx=dtx21x6dx=13dt1t2=13[12log|1+t1t|]+C 
Q7. Integrate the function : 
Answer. x1x21dx=xx21dx1x21dx....(1)  For xx21dx, let x21=t2xdx=dtxx21dx=12dtt =12t12dt=12[2t12]=t=x21 From ( 1), we obtain  x1x21dx=xx21dx1x21dx[1x2a2dt=log|x+x2a2|] 
Q8. Integrate the function : x2x6+a6

Answer.  Let x3=t3x2dx=dtx2x6+a6dx=13dtt2+(a3)2 
Q9. Integrate the function : sec2xtan2x+4

Answer.   Let tan x=tsec2xdx=dt 
Q10. Integrate the function : 1x2+2x+2

Answer.  1x2+2x+2dx=1(x+1)2+(1)2dx Let x+1=tdx=dt1x2+2x+2dx=1t2+1dt 

Q11. Integrate the function : 19x2+6x+5

Answer.  19x2+6x+5dx=1(3x+1)2+(2)2dx Let (3x+1)=t3dx=dt1(3x+1)2+(2)2dx=131t2+22dt 
Q12. Integrate the function : 176xx2

Answer.  76xx2 can be written as 7(x2+6x+99) Therefore, 7(x2+6x+99)=16(x2+6x+9)=16(x+3)2 =(4)2(x+3)2176xx2dx=1(4)2(x+3)2dx Let x+3=tdx=dt1(4)2(x+3)2dx=1(4)2(t)2dt 
Q13. Integrate the function : 
Answer.  (x1)(x2) can be written as x23x+2 Therefore, x23x+2=x23x+9494+2=(x32)214=(x32)2(12)2 <