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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 49  Maths Textbook Solution.

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Answer: \frac{x^{3}}{3}\tan ^{-1}x-\frac{x^{2}}{6}+\frac{1}{6}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 x^{2}\tan ^{-1}xdx


            \int x^{2}\tan ^{-1}xdx

           \begin{aligned} &=\tan ^{-1} x \int x^{2} d x-\int \frac{d}{d x} \tan ^{-1} x\left(\int x^{2} d x\right) d x \\ &=\frac{x^{3}}{3} \tan ^{-1} x-\frac{1}{3} \int \frac{x \cdot x^{2}}{1+x^{2}} d x \\ &=\frac{x^{3}}{3} \tan ^{-1} x-\frac{1}{3} \int \frac{x\left(1+x^{2}\right)}{1+x^{2}} d x+\frac{1}{6} \int \frac{2 x}{1+x^{2}} d x \\ &=\frac{x^{3}}{3} \tan ^{-1} x-\frac{x^{2}}{6}+\frac{1}{6} \log \left(1+x^{2}\right)+c \end{aligned}

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