NCERT Solutions Class 12 Maths (integrals) Miscellaneous Exercise

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-7(integrals) Miscellaneous Exercise 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.

Q1. Integrate the function

Answer. $\frac{1}{x-{x}^{3}}=\frac{1}{x\left(1-{x}^{2}\right)}=\frac{1}{x\left(1-x\right)\left(1+x\right)}$  Equating the coefficients of ${x}^{2}$ , x, and constant term, we obtain $\begin{array}{l}-A+B-C=0\\ B+C=0\\ A=1\end{array}$ On solving these equations, we obtain  From equation (1), we obtain $\frac{1}{x\left(1-x\right)\left(1+x\right)}=\frac{1}{x}+\frac{1}{2\left(1-x\right)}-\frac{1}{2\left(1+x\right)}$ $\begin{array}{l}\frac{1}{x\left(1-x\right)\left(1+x\right)}=\frac{1}{x}+\frac{1}{2\left(1-x\right)}-\frac{1}{2\left(1+x\right)}\\ ⇒\int \frac{1}{x\left(1-x\right)\left(1+x\right)}dx=\int \frac{1}{x}dx+\frac{1}{2}\int \frac{1}{1-x}dx-\frac{1}{2}\int \frac{1}{1+x}dx\end{array}$

Q2. Integrate the function

Answer. $\begin{array}{rl}\frac{1}{\sqrt{x+a}+\sqrt{x+b}}& =\frac{1}{\sqrt{x+a}+\sqrt{x+b}}×\frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}\\ & =\frac{\sqrt{x+a}-\sqrt{x+b}}{\left(x+a\right)-\left(x+b\right)}\\ & =\frac{\left(\sqrt{x+a}-\sqrt{x+b}\right)}{a-b}\end{array}$

Q3. Integrate the function

Answer.  $\begin{array}{rl}⇒\int \frac{1}{x\sqrt{ax-{x}^{2}}}dx& =\int \frac{1}{\frac{a}{t}\sqrt{a\cdot \frac{a}{t}-{\left(\frac{a}{t}\right)}^{2}}}\left(-\frac{a}{{t}^{2}}dt\right)\\ & =-\int \frac{1}{at}\cdot \frac{1}{\sqrt{\frac{1}{t}-\frac{1}{{t}^{2}}}}dt\end{array}$ $\begin{array}{rl}& =-\frac{1}{a}\int \frac{1}{\sqrt{\frac{{t}^{2}}{t}-\frac{{t}^{2}}{{t}^{2}}}}dt\\ & =-\frac{1}{a}\int \frac{1}{\sqrt{t-1}}dt\\ & =-\frac{1}{a}\left[2\sqrt{t-1}\right]+\mathrm{C}\\ & =-\frac{1}{a}\left[2\sqrt{\frac{a}{x}-1}\right]+\mathrm{C}\end{array}$

Q4. Integrate the function

Answer.  $\begin{array}{rl}& =\frac{{\left({x}^{4}+1\right)}^{\frac{-3}{4}}}{{x}^{5}\cdot {\left({x}^{4}\right)}^{\frac{3}{4}}}\\ & =\frac{1}{{x}^{5}}{\left(\frac{{x}^{4}+1}{{x}^{4}}\right)}^{\frac{3}{4}}\\ =& \frac{1}{{x}^{5}}{\left(1+\frac{1}{{x}^{4}}\right)}^{-\frac{3}{4}}\end{array}$  $\begin{array}{rl}& =-\frac{1}{4}\int \left(1+t{\right)}^{\frac{3}{4}}dt\\ & =-\frac{1}{4}\left[\frac{\left(1+t{\right)}^{\frac{1}{4}}}{\frac{1}{4}}\right]+C\end{array}$

Q5. Integrate the function

Answer.  $\begin{array}{rl}\therefore \int \frac{1}{{x}^{\frac{1}{2}}+{x}^{\frac{1}{3}}}dx& =\int \frac{1}{{x}^{\frac{1}{3}}\left(1+{x}^{\frac{1}{6}}\right)}dx\\ & =\int \frac{6{t}^{5}}{{t}^{2}\left(1+t\right)}dt\\ & =6\int \frac{{t}^{3}}{\left(1+t\right)}dt\end{array}$

Q6. Integrate the function