Q55 / 150 · MHT CET 2017 · Mathematics

$\int_{0}^{1} x \tan^{- 1} x d x =$

A
$\frac{π}{4} + \frac{1}{2}$
Correct answer
B
$\frac{π}{4} - \frac{1}{2}$
C
$\frac{1}{2} - \frac{π}{4}$
D
$- \frac{π}{4} - \frac{1}{2}$

Solution

\int_{\begin{matrix} 0 \\ (u v r u l e) \end{matrix}}^{1} x \tan^{- 1} x d x = {[\tan^{- 1} x \int x d x]}_{0}^{1} - \int_{0}^{1} (\frac{d}{d x} \tan^{- 1} x \int x d x) d x

= {(\tan^{- 1} x \cdot \frac{x^{2}}{2})}_{0}^{1} - \int_{0}^{1} \frac{1}{1 + x^{2}} \cdot \frac{x^{2}}{2} d x

= (\frac{π}{4} \cdot \frac{1}{2} - 0) - \frac{1}{2} \int_{0}^{1} \frac{x^{2}}{1 + x^{2}} d x

= \frac{π}{8} - \frac{1}{2} \int_{0}^{1} (1 - \frac{1}{1 + x^{2}}) d x

= \frac{π}{8} - \frac{1}{2} {[x - \tan^{- 1} x]}_{0}^{1}

= \frac{π}{8} - \frac{1}{2} [(1 - 0) - (\frac{π}{4} - 0)] = \frac{π}{8} - \frac{1}{2} + \frac{π}{8} = \frac{π}{4} - \frac{1}{2}

Hence, the correct answer is option: 'B'.

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