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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.

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

Answers (1)

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

Solution:

            \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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