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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 36 Maths Textbook Solution.

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

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

Hint: 2\tan ^{-1}x=\sin ^{-1}x\frac{2x}{1+x^{2}}


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


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