(x + 1) ^2 , ,dx x( x^2 + 1) is equal to - Vidyadip

MCQ

$\int {\frac{{{{(x + 1)}^2}\,\,dx}}{{x({x^2} + 1)}}} $ is equal to

A
${\log _e}x + c$
✓
${\log _e}x + 2{\tan ^{ - 1}}x + c$
C
${\log _e}\frac{1}{{{x^2} + 1}} + c$
D
${\log _e}\{ x({x^2} + 1)\} + c$

✓

Answer

Correct option: B.

${\log _e}x + 2{\tan ^{ - 1}}x + c$

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

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