_ ^ x^2 + 1 x^4 - x^2 + 1 ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{{x^2} + 1}}{{{x^4} - {x^2} + 1}}\;dx = } $

A
${\tan ^{ - 1}}\left( {\frac{{1 + {x^2}}}{x}} \right) + c$
B
${\cot ^{ - 1}}\left( {\frac{{1 + {x^2}}}{x}} \right) + c$
✓
${\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{x}} \right) + c$
D
${\cot ^{ - 1}}\left( {\frac{{{x^2} - 1}}{x}} \right) + c$

✓

Answer

Correct option: C.

${\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{x}} \right) + c$

c
(c)$\int_{}^{} {\frac{{{x^2} + 1}}{{{x^4} - {x^2} + 1}}\,dx = \int_{}^{} {\frac{{\left( {1 + \frac{1}{{{x^2}}}} \right)}}{{{x^2} + \frac{1}{{{x^2}}} - 1}}} } $
$ = \int_{}^{} {\frac{{1 + \frac{1}{{{x^2}}}}}{{{{\left( {x - \frac{1}{x}} \right)}^2} + 1}}\,dx = \int_{}^{} {\frac{{dt}}{{{t^2} + 1}}} = {{\tan }^{ - 1}}t + c} $
$\left\{ {{\rm{Putting}}\,x - \frac{1}{x} = t \Rightarrow \left( {1 + \frac{1}{{{x^2}}}} \right)\,dx = dt} \right\}$
$ = {\tan ^{ - 1}}\left( {x - \frac{1}{x}} \right) + c = {\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{x}} \right) + c.$

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