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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Mathematics
Chapter
Definite Integration

BITSAT 2009 · Mathematics · Definite Integration

Evaluate: ( ^ -1 x d x)

Evaluate: (\intx \tan ^{-1} x d x)

Options

  1. A(\frac{1}{2}\left(x^2+1\right) \tan ^{-1} x-12\frac{1}{2}x+c)
  2. B(\frac{1}{2}\left(x^2+1\right) \tan ^{-1} x+12\frac{1}{2}x+c)
  3. C(\frac{1}{2}\left(\mathrm{x}^2-1\right) \tan ^{-1} x−12x+c\mathrm{x}-\frac{1}{2} \mathrm{x}+\mathrm{c})
  4. DNone of these

Correct answer

A. (\frac{1}{2}\left(x^2+1\right) \tan ^{-1} x-12\frac{1}{2}x+c)

Step-by-step solution

∫xtan⁡−1xdx=x22tan⁡−1x−12∫x21+x2dx=x22tan⁡−1x−12∫\lleft[1−11+x2]dx=12(x2+1)tan⁡−1x−12x+c\begin{aligned}& \int x \tan ^{-1} x d x=\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int \frac{x^2}{1+x²}d x \\ & =\frac{x^2}{2} \tan ^{-1}x-\frac{1}{2} \int \lleft[1-\frac{1}{1+x²}]d x \\ & =\frac{1}{2}\left(x^2+1\right)\tan ^{-1}x-\frac{1}{2} x+c \end{aligned})$
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