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STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
MCQ
For $x > 1,\;\int_{}^{} {\frac{1}{{x({x^4} - 1)}}\;dx = } $
  • A
    $\log \frac{{{x^4} - 1}}{{{x^4}}} + K$
  • ✓
    $\frac{1}{4}\log \frac{{{x^4} - 1}}{{{x^4}}} + K$
  • C
    $\log \frac{{{x^4} - 1}}{x} + K$
  • D
    $\frac{1}{4}\log \frac{{{x^4} - 1}}{x} + K$
✓

Answer

Correct option: B.
$\frac{1}{4}\log \frac{{{x^4} - 1}}{{{x^4}}} + K$
b
(b)$\int_{}^{} {\frac{1}{{x({x^4} - 1)}}\,dx = \frac{1}{4}\int_{}^{} {\left[ {\frac{{4{x^3}}}{{({x^4} - 1)}} - \frac{4}{x}} \right]\,dx} } $
$ = \frac{1}{4}[\log ({x^4} - 1) - 4\log x] + c = \frac{1}{4}\log \frac{{{x^4} - 1}}{{{x^4}}} + c$.

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