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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 108 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 108 Maths Textbook Solution.

Answers (1)

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

Hint: in this equation we will us ILATE method and then differentiate the terms

Given: x^{2} \tan ^{-1} x d x

Solution: Considering \tan ^{-1} x as first function and x^{2} as second

 

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