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
- A(\frac{1}{2}\left(x^2+1\right) \tan ^{-1} x-21x+c)
- B(\frac{1}{2}\left(x^2+1\right) \tan ^{-1} x+21x+c)
- C(\frac{1}{2}\left(\mathrm{x}^2-1\right) \tan ^{-1} x−21x+c)
- DNone of these
Correct answer
A. (\frac{1}{2}\left(x^2+1\right) \tan ^{-1} x-21x+c)
Step-by-step solution
∫xtan−1xdx=2x2tan−1x−21∫1+x2x2dx=2x2tan−1x−21∫\lleft[1−1+x21]dx=21(x2+1)tan−1x−21x+c)$
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