#### Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 17.

Answer:$\frac{\pi}{2}-\log 2$

Hint: We use indefinite integral formula then put limits to solve this integral.

Given: $\int_{0}^{1}\tan ^{-1}\left ( \frac{2x}{1-x^2} \right )dx$

Solution: $I=\int_{0}^{1}\tan ^{-1}\left ( \frac{2x}{1-x^2} \right )dx$

Put $x=\tan \theta$

$dx=\sec^2\theta \; d\theta$

When x=0  then $\theta=0$  and

when x=1  then $\theta=\frac{\pi}{4}$

\begin{aligned} &I=\int_{0}^{\frac{\pi}{4}} \tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right) \sec ^{2} \theta d \theta\; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right] \\ &=\int_{0}^{\frac{\pi}{4}} \tan ^{-1}(\tan 2 \theta) \sec ^{2} \theta d \theta \\ &=\int_{0}^{\frac{\pi}{4}} 2 \theta \sec ^{2} \theta d \theta \\ &=2 \int_{0}^{\frac{\pi}{4}} \theta \sec ^{2} \theta d \theta \end{aligned}

Applying integration by parts method, then

\begin{aligned} &=2\left\{\left[\theta \int \sec ^{2} \theta d \theta\right]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\left[\frac{d(\theta)}{d \theta} \int \sec ^{2} \theta d \theta\right] d \theta\right\} \\ &=2\left\{[\theta \tan \theta]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}} 1 \tan \theta d \theta\right\} \quad\left[\int \sec ^{2} x d x=\tan x, \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right] \\ &=2\left\{\left[\frac{\pi}{4} \tan \frac{\pi}{4}-0 \times \tan 0\right]-\int_{0}^{\frac{\pi}{4}} \tan \theta d \theta\right\} \quad\left[\int \tan x d x=-\log |\cos x|, \tan \frac{\pi}{4}=1, \tan 0=0\right] \end{aligned}

\begin{aligned} &=2\left\{\left[\frac{\pi}{4} \cdot 1-0\right]-[-\log |\cos \theta|]_{0}^{\frac{\pi}{4}}\right\} \\ &=\frac{2 \pi}{4}+2\left[\log \frac{1}{\sqrt{2}}-\log 1\right] \; \; \; \; \; \; \; \: \: \: \: \quad\left[\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \cos 0=1\right] \\ &=\frac{\pi}{2}+2\left[\log \frac{1}{\sqrt{2}}-0\right] \: \: \: \: \: \: \: \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\log 1=0] \end{aligned}

\begin{aligned} &=\frac{\pi}{2}+2 \log (2)^{-\frac{1}{2}} \\ &=\frac{\pi}{2}+\left(-\frac{1}{2}\right) \times 2 \log 2 \quad\left[\log a^{m}=m \log a\right] \\ &=\frac{\pi}{2}-\log 2 \end{aligned}