x^2 x^2 + 4 , ,dx equals to - Vidyadip

MCQ

$\int {\frac{{{x^2}}}{{{x^2} + 4}}\,\,dx} $ equals to

✓
$x - 2{\tan ^{ - 1}}(x/2) + c$
B
$x + 2{\tan ^{ - 1}}(x/2) + c$
C
$x - 4{\tan ^{ - 1}}(x/2) + c$
D
$x + 4{\tan ^{ - 1}}(x/2) + c$

✓

Answer

Correct option: A.

$x - 2{\tan ^{ - 1}}(x/2) + c$

a
(a) $I = \int {\frac{{{x^2}}}{{{x^2} + 4}}dx} $$ = \int {\frac{{{x^2} + 4 - 4}}{{({x^2} + 4)}}dx} $
==> $I = \int {\left[ {1 - \frac{4}{{{x^2} + 4}}} \right]\,dx} $$ = \int {dx - \int {\frac{4}{{{x^2} + 4}}dx} } $
$ \Rightarrow I = x - 4\int {\frac{{dx}}{{{x^2} + {{(2)}^2}}}} $$ = x - \frac{4}{2}{\tan ^{ - 1}}(x/2) + c$
$ = x - 2{\tan ^{ - 1}}\frac{x}{2} + c$.

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