_ ^ dx x[ ( x) ^2 + 4 x - 1] = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{x[{{(\log x)}^2} + 4\log x - 1]}}} = $

✓
$\frac{1}{{2\sqrt 5 }}\log \left[ {\frac{{\log x + 2 - \sqrt 5 }}{{\log x + 2 + \sqrt 5 }}} \right] + c$
B
$\frac{1}{{\sqrt 5 }}\log \left[ {\frac{{\log x + 2 - \sqrt 5 }}{{\log x + 2 + \sqrt 5 }}} \right] + c$
C
$\frac{1}{{2\sqrt 5 }}\log \left[ {\frac{{\log x + 2 + \sqrt 5 }}{{\log x + 2 - \sqrt 5 }}} \right] + c$
D
$\frac{1}{{\sqrt 5 }}\log \left[ {\frac{{\log x + 2 + \sqrt 5 }}{{\log x + 2 - \sqrt 5 }}} \right] + c$

✓

Answer

Correct option: A.

$\frac{1}{{2\sqrt 5 }}\log \left[ {\frac{{\log x + 2 - \sqrt 5 }}{{\log x + 2 + \sqrt 5 }}} \right] + c$

a
(a) Put $\log x = t \Rightarrow \frac{1}{x}\,dx = dt,$ then
$\int_{}^{} {\frac{{dx}}{{x[{{(\log x)}^2} + 4\log x - 1]}}} = \int_{}^{} {\frac{{dt}}{{{t^2} + 4t - 1}}} $
$ = \int_{}^{} {\frac{{dt}}{{{{(t + 2)}^2} - {{(\sqrt 5 )}^2}}} = \frac{1}{{2\sqrt 5 }}\log \left[ {\frac{{t + 2 - \sqrt 5 }}{{t + 2 + \sqrt 5 }}} \right]} $
$ = \frac{1}{{2\sqrt 5 }}\log \left[ {\frac{{\log x + 2 - \sqrt 5 }}{{\log x + 2 + \sqrt 5 }}} \right] + c$.

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