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

MCQ

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

A
$\frac{1}{2}\log \left( {\frac{{{x^2} + x + 1}}{{{x^2} - x + 1}}} \right) + c$
B
$\frac{1}{2}\log \left( {\frac{{{x^2} - x - 1}}{{{x^2} + x + 1}}} \right) + c$
C
$\log \left( {\frac{{{x^2} - x + 1}}{{{x^2} + x + 1}}} \right) + c$
✓
$\frac{1}{2}\log \left( {\frac{{{x^2} - x + 1}}{{{x^2} + x + 1}}} \right) + c$

✓

Answer

Correct option: D.

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

d
(d) The given function can be written as

$\int_{}^{} {\frac{{\left( {1 - \frac{1}{{{x^2}}}} \right)}}{{{{\left( {x + \frac{1}{x}} \right)}^2} - 1}}} \,dx$
Put $x + \frac{1}{x} = t \Rightarrow \left( {1 - \frac{1}{{{x^2}}}} \right)\,dx = dt,$ then it reduces to
$\int_{}^{} {\frac{{dt}}{{{t^2} - 1}}} = \frac{1}{2}\log \left| {\frac{{t - 1}}{{t + 1}}} \right| + c$
$ = \frac{1}{2}\log \left( {\frac{{x + \frac{1}{x} - 1}}{{x + \frac{1}{x} + 1}}} \right) + c = \frac{1}{2}\log \left( {\frac{{{x^2} - x + 1}}{{{x^2} + x + 1}}} \right) + c.$

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