_ ^ x^2 + 1 x( x^2 - 1) ;dx is equal to - Vidyadip

MCQ

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

✓
$\log \frac{{{x^2} - 1}}{x} + c$
B
$ - \log \frac{{{x^2} - 1}}{x} + c$
C
$\log \frac{x}{{{x^2} + 1}} + c$
D
$ - \log \frac{x}{{{x^2} + 1}} + c$

✓

Answer

Correct option: A.

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

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

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