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Need Solution for R.D. Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 109 Maths Textbook Solution.

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

Hint: to solve this equation we have to use Byparts method

Given:\int \tan ^{-1} \sqrt{x} d x

Solution: \text { let } \sqrt{x}=t

\frac{1}{2 \sqrt{x}} d x=d t

d x=2 t d t

I=\int \tan ^{-1} t .2 t d t

I=\tan ^{-1} t . t^{2}-\int \frac{t^{2}+1-1}{1+t^{2}} d t \ldots . \text { using Byparts }

I=\tan ^{-1} t \cdot t^{2}-\int d t+\int \frac{1}{1+t^{2}}

I=\tan ^{-1} t \cdot t^{2}-t+\tan ^{-1} t+C

I=\tan ^{-1} t\left(1+t^{2}\right)-t+C

I=\tan ^{-1} \sqrt{x}(1+x)-\sqrt{x}+C


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