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7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
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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