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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{x}{{{x^4} - 1}}dx = } $
  • ✓
    $\frac{1}{4}\log \left[ {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \right] + c$
  • B
    $\frac{1}{4}\log \left[ {\frac{{{x^2} + 1}}{{{x^2} - 1}}} \right] + c$
  • C
    $\frac{1}{2}\log \left[ {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \right] + c$
  • D
    $\frac{1}{2}\log \left[ {\frac{{{x^2} + 1}}{{{x^2} - 1}}} \right] + c$
✓

Answer

Correct option: A.
$\frac{1}{4}\log \left[ {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \right] + c$
a
(a)$\int_{}^{} {\frac{x}{{({x^4} - 1)}}\,dx = \frac{1}{2}\int_{}^{} {\left[ {\frac{x}{{{x^2} - 1}} - \frac{x}{{{x^2} + 1}}} \right]} \,dx} $
$ = \frac{1}{4}\log ({x^2} - 1) - \frac{1}{4}\log ({x^2} + 1) = \frac{1}{4}\log \left( {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \right) + c$.
Aliter : Put $t = {x^2} \Rightarrow dt = 2x\,dx,$ then
$\int_{}^{} {\frac{x}{{{x^4} - 1}}\,dx} = \frac{1}{2}\int_{}^{} {\frac{{dt}}{{{t^2} - 1}} = \frac{1}{2}.\frac{1}{2}\log \frac{{t - 1}}{{t + 1}} + c} $
$ = \frac{1}{4}\log \left[ {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \right] + c.$

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