_ ^ dx 2 x^2 + x + 1 ;equals - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{2{x^2} + x + 1}}} \;$equals

A
$\frac{1}{{\sqrt 7 }}{\tan ^{ - 1}}\left( {\frac{{4x + 1}}{{\sqrt 7 }}} \right) + c$
B
$\frac{1}{{2\sqrt 7 }}{\tan ^{ - 1}}\left( {\frac{{4x + 1}}{{\sqrt 7 }}} \right) + c$
C
$\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{4x + 1}}{{\sqrt 7 }}} \right) + c$
✓
None of these

✓

Answer

Correct option: D.

None of these

d
(d)$I = \int_{}^{} {\frac{{dx}}{{2{x^2} + x + 1}}} = \int_{}^{} {\frac{{dx}}{{2\left( {{x^2} + \frac{x}{2} + \frac{1}{2}} \right)}}} $
$ = \frac{1}{2}\int_{}^{} {\frac{{dx}}{{{x^2} + \frac{x}{2} + \frac{1}{{16}} - \frac{1}{{16}} + \frac{1}{2}}}} = \frac{1}{2}\int_{}^{} {\frac{{dx}}{{{{\left( {x + \frac{1}{4}} \right)}^2} + {{\left( {\frac{{\sqrt 7 }}{4}} \right)}^2}}}} $
$ = \frac{1}{2}\frac{1}{{\frac{{\sqrt 7 }}{4}}}{\tan ^{ - 1}}\frac{{[x + (14)]}}{{\sqrt 7 4}}$$ = \frac{2}{{\sqrt 7 }}{\tan ^{ - 1}}\frac{{(4x + 1)}}{{\sqrt 7 }} + C$.

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