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

MCQ

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

A
$\frac{1}{3}{\tan ^{ - 1}}\left( {\frac{{2{x^2} + 1}}{3}} \right)$
✓
$\frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{2{x^2} + 1}}{{\sqrt 3 }}} \right)$
C
$\frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}(2{x^2} + 1)$
D
None of these

✓

Answer

Correct option: B.

$\frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{2{x^2} + 1}}{{\sqrt 3 }}} \right)$

b
(b)$I = \int_{}^{} {\frac{x}{{{x^4} + {x^2} + 1}}\,dx = \int_{}^{} {\frac{{xdx}}{{({x^2} + x + 1)\,({x^2} - x + 1)}}} } $
$I = \frac{1}{2}\int_{}^{} {\frac{{dx}}{{{x^2} - x + 1}} - \frac{1}{2}\int_{}^{} {\frac{{dx}}{{{x^2} + x + 1}}} } $
$I = \frac{1}{2}\int_{}^{} {\frac{{dx}}{{{{\left( {x - \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} - \frac{1}{2}\int_{}^{} {\frac{{dx}}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} } $
$I = \frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{x - \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}}} \right) - \frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{x + \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}}} \right)$
$I = \frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{2{x^2} + 1}}{{\sqrt 3 }}} \right)$.

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