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CUET · MATHS · PYQ PAPER 2023

\(\int x^2 \tan ^{-1} x d x\)=

  1. A \(\frac{x^3}{3} \tan ^{-1} x-\frac{x^2}{6}+\frac{1}{6} \log \left|x^2+1\right|+C\)
  2. B \(\frac{x^3}{3} \tan ^{-1} x+\frac{x^2}{6}+\frac{1}{6} \log \left|x^2+1\right|+C\)
  3. C \(\frac{x^3}{3} \tan ^{-1} x-\frac{x^2}{6}-\frac{1}{6} \log \left|x^2+1\right|+C\)
  4. D \(\frac{x^3}{3} \tan ^{-1} x+\frac{x^2}{6}-\frac{1}{6} \log \left|x^2+1\right|+C\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{x^3}{3} \tan ^{-1} x-\frac{x^2}{6}+\frac{1}{6} \log \left|x^2+1\right|+C\)

Step-by-step Solution

Detailed explanation

\(\int x^2 \tan ^{-1} x d x = \tan ^{-1} x \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{1}{1+x^2} d x\) \(= \frac{x^3}{3} \tan ^{-1} x - \frac{1}{3} \int \frac{x^3}{1+x^2} d x\) \(= \frac{x^3}{3} \tan ^{-1} x - \frac{1}{3} \int \left(x - \frac{x}{1+x^2}\right) d x\)…
From CUET
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