_ ^ x^2 - 1 x^4 + x^2 + 1 dx is equal to - Vidyadip

MCQ

$\int_{}^{} {\frac{{{x^2} - 1}}{{{x^4} + {x^2} + 1}}dx} $ is equal to

A
$\log ({x^4} + {x^2} + 1) + c$
✓
$\frac{1}{2}\log \frac{{{x^2} - x + 1}}{{{x^2} + x + 1}} + c$
C
$\frac{1}{2}\log \frac{{{x^2} + x + 1}}{{{x^2} - x + 1}} + c$
D
$\log \frac{{{x^2} - x + 1}}{{x + x + 1}} + c$

✓

Answer

Correct option: B.

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

b
(b) $I = \int_{}^{} {\frac{{{x^2} - 1}}{{{x^4} + {x^2} + 1}}} \,dx$$ = \int_{}^{} {\frac{{{x^2}\left( {1 - \frac{1}{{{x^2}}}} \right)}}{{{x^2}\left[ {{{\left( {x + \frac{1}{x}} \right)}^2} - 1} \right]}}\,dx} $
Put $\left( {x + \frac{1}{x}} \right) = t \Rightarrow \left( {1 - \frac{1}{{{x^2}}}} \right)\,dx = dt$
$I = \int_{}^{} {\frac{{dt}}{{{t^2} - 1}}} = \frac{1}{2}\log \left| {\,\frac{{t - 1}}{{t + 1}}\,} \right| + c$
$\therefore $ $I = \frac{1}{2}\log \,\left| {\frac{{{x^2} - x + 1}}{{{x^2} + x + 1}}} \right| + c$ ==> $a = \frac{1}{2},\,b = \frac{1}{2}$.

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