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need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 15

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Answer: 2 \sqrt{\tan ^{-1} x}+c

Hint: Use substitution method to solve this integral.

Given: \int \frac{1}{\sqrt{\tan ^{-1} x}\left(1+x^{2}\right)} d x


        \text { Let } I=\int \frac{1}{\sqrt{\tan ^{-1} x}\left(1+x^{2}\right)} d x

        \text { Put } \tan ^{-1} x=t \Rightarrow \frac{1}{1+x^{2}} d x=d t \Rightarrow d x=\left(1+x^{2}\right) d t \text { then }

        \begin{aligned} I &=\int \frac{1}{\sqrt{t}\left(1+x^{2}\right)}\left(1+x^{2}\right) d t=\int \frac{1}{\sqrt{t}} d t \\ &=\int t^{\frac{-1}{2}} d t=\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \end{aligned}

            \begin{aligned} &=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\mathrm{c}=2 \sqrt{t}+c \\ &=2 \sqrt{\tan ^{-1} x}+c\left[\because t=\tan ^{-1} x\right] \end{aligned}

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