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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 45 Maths Textbook Solution.

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Answer: \frac{x^{3}}{3}\tan ^{-1}x-\frac{x^{2}}{6}+\frac{1}{6}log|1+x^{2}|+c

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

Given:\int x^{2}\tan ^{-1}xdx


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