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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{1}{{x - {x^3}}}\;dx = } $
  • A
    $\frac{1}{2}\log \frac{{(1 - {x^2})}}{{{x^2}}} + c$
  • B
    $\log \frac{{(1 - x)}}{{x(1 + x)}} + c$
  • C
    $\log x(1 - {x^2}) + c$
  • ✓
    $\frac{1}{2}\log \frac{{{x^2}}}{{(1 - {x^2})}} + c$
✓

Answer

Correct option: D.
$\frac{1}{2}\log \frac{{{x^2}}}{{(1 - {x^2})}} + c$
d
(d)$\int_{}^{} {\frac{1}{{x - {x^3}}}\,dx = \int_{}^{} {\frac{1}{{x(1 + x)(1 - x)}}\,dx} } $
$ = \frac{1}{2}\int_{}^{} {\left( {\frac{2}{x} - \frac{1}{{1 + x}} + \frac{1}{{1 - x}}} \right)\,dx} $
$ = \frac{1}{2}[2\log x - \log (1 + x) - \log (1 - x)] = \frac{1}{2}\log \frac{{{x^2}}}{{(1 - {x^2})}} + c$.

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