_ ^ dx ( x^2 + 1)( x^2 + 4) = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{({x^2} + 1)({x^2} + 4)}} = } $

A
$\frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{3}{\tan ^{ - 1}}\frac{x}{2} + c$
B
$\frac{1}{3}{\tan ^{ - 1}}x + \frac{1}{3}{\tan ^{ - 1}}\frac{x}{2} + c$
✓
$\frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{6}{\tan ^{ - 1}}\frac{x}{2} + c$
D
${\tan ^{ - 1}}x - 2{\tan ^{ - 1}}\frac{x}{2} + c$

✓

Answer

Correct option: C.

$\frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{6}{\tan ^{ - 1}}\frac{x}{2} + c$

c
(c)$\int_{}^{} {\frac{{dx}}{{({x^2} + 1)({x^2} + 4)}}} = \frac{1}{3}\left[ {\int_{}^{} {\frac{{dx}}{{{x^2} + 1}} - \int_{}^{} {\frac{{dx}}{{{x^2} + 4}}} } } \right]$
$ = \frac{1}{3}\left[ {{{\tan }^{ - 1}}x - \frac{1}{2}{{\tan }^{ - 1}}\frac{x}{2}} \right] + c = \frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{6}{\tan ^{ - 1}}\frac{x}{2} + c$.

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