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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}}{{({x^2} + 1)({x^2} + 4)}} = } $
  • A
    $\frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{3}{\tan ^{ - 1}}\frac{x}{2} + c$
  • B
    $\frac{1}{3}{\tan ^{ - 1}}x + \frac{1}{3}{\tan ^{ - 1}}\frac{x}{2} + c$
  • ✓
    $\frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{6}{\tan ^{ - 1}}\frac{x}{2} + c$
  • D
    ${\tan ^{ - 1}}x - 2{\tan ^{ - 1}}\frac{x}{2} + c$
✓

Answer

Correct option: C.
$\frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{6}{\tan ^{ - 1}}\frac{x}{2} + c$
c
(c)$\int_{}^{} {\frac{{dx}}{{({x^2} + 1)({x^2} + 4)}}} = \frac{1}{3}\left[ {\int_{}^{} {\frac{{dx}}{{{x^2} + 1}} - \int_{}^{} {\frac{{dx}}{{{x^2} + 4}}} } } \right]$
$ = \frac{1}{3}\left[ {{{\tan }^{ - 1}}x - \frac{1}{2}{{\tan }^{ - 1}}\frac{x}{2}} \right] + c = \frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{6}{\tan ^{ - 1}}\frac{x}{2} + c$.

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