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Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 43 Maths Textbook Solution.

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Answer: 2x\tan ^{-1}x-log\left ( 1+x^{2} \right )+c

Hint: \int uvdx=u\int vdx-\int \left ( \frac{d}{dx}u\int vdx \right )dx

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


           \begin{aligned} &I=2 \int \tan ^{-1} x d x \\ &=2\left[\tan ^{-1} x \int d x-\int\left[\frac{d\left(\tan ^{-1} x\right)}{d x} \int d x\right] d x\right] \\ &=2\left[x \tan ^{-1} x-\int \frac{x}{1+x^{2}} d x\right] \end{aligned}

\int \frac{x}{1+x^{2}}dx

Put 1+x^{2}=t\Rightarrow 2xdx=dt\Rightarrow xdx=\frac{dt}{2}

                                \begin{aligned} &\Rightarrow \int \frac{x}{1+x^{2}} d x=\frac{1}{2} \int \frac{d t}{t} \\ &=\frac{1}{2} \log t+c \\ &=\frac{1}{2} \log \left(1+x^{2}\right)+c \end{aligned}

2\left[x \tan ^{-1} x-\int \frac{x}{1+x^{2}} d x\right]=2 x \tan ^{-1} x-\log \left(1+x^{2}\right)+c

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