Get Answers to all your Questions

header-bg qa

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}


Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support