_ ^ 1 x^2 ( x^2 + a^2 )dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{1}{{{x^2}}}\log ({x^2} + {a^2})dx = } $

A
$\frac{1}{x}\log ({x^2} + {a^2}) + \frac{2}{a}{\tan ^{ - 1}}\frac{x}{a} + c$
✓
$ - \frac{1}{x}\log ({x^2} + {a^2}) + \frac{2}{a}{\tan ^{ - 1}}\frac{x}{a} + c$
C
$ - \frac{1}{x}\log ({x^2} + {a^2}) - \frac{2}{a}{\tan ^{ - 1}}\frac{x}{a} + c$
D
None of these

✓

Answer

Correct option: B.

$ - \frac{1}{x}\log ({x^2} + {a^2}) + \frac{2}{a}{\tan ^{ - 1}}\frac{x}{a} + c$

b
(b) $\int_{}^{} {\frac{1}{{{x^2}}}\log ({x^2} + {a^2})\,dx} = \int_{}^{} {{x^{ - 2}}\log ({x^2} + {a^2})\,dx} $
$ = \frac{{ - \log ({x^2} + {a^2})}}{x} + \int_{}^{} {\frac{{2x}}{{({x^2} + {a^2})}}.\frac{1}{x} + c} $
$ = \frac{{ - \log ({x^2} + {a^2})}}{x} + \frac{2}{a}{\tan ^{ - 1}}\frac{x}{a} + c.$

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