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

MCQ

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

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

✓

Answer

Correct option: D.

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

d
(d)$\int_{}^{} {\frac{{{x^2} + 1}}{{{x^4} + 1}}} \,dx = \int_{}^{} {\frac{{\left( {1 + \frac{1}{{{x^2}}}} \right)}}{{\left( {{x^2} + \frac{1}{{{x^2}}}} \right)}}} \,dx = \int_{}^{} {\frac{{\left( {1 + \frac{1}{{{x^2}}}} \right)\,dx}}{{{{\left( {x - \frac{1}{x}} \right)}^2} + 2}}} $
Put $x - \frac{1}{x} = t \Rightarrow \left( {1 + \frac{1}{{{x^2}}}} \right)\,dx = dt,$ then the required integral is $\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{{\sqrt 2 x}}} \right) + c$.

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