_ ^ x ;dx x^3 = - Vidyadip

MCQ

$\int_{}^{} {\frac{{\log x\;dx}}{{{x^3}}} = } $

A
$\frac{1}{{4{x^2}}}(2\log x - 1) + c$
✓
$ - \frac{1}{{4{x^2}}}(2\log x + 1) + c$
C
$\frac{1}{{4{x^2}}}(2\log x + 1) + c$
D
$\frac{1}{{4{x^2}}}(1 - 2\log x) + c$

✓

Answer

Correct option: B.

$ - \frac{1}{{4{x^2}}}(2\log x + 1) + c$

b
(b)$\int_{}^{} {\frac{{\log x}}{{{x^3}}}dx = \int_{}^{} {{x^{ - 3}}\log x\;dx} } $
$ = - \frac{{\log x}}{{2{x^2}}} + \int_{}^{} {\frac{1}{x}.\frac{1}{{2{x^2}}} + c = - \frac{{\log x}}{{2{x^2}}} + \frac{1}{2}.\frac{{{x^{ - 2}}}}{{ - 2}} + c} $
$ = - \frac{{\log x}}{{2{x^2}}} - \frac{1}{{4{x^2}}} + c = - \frac{1}{{4{x^2}}}(2\log x + 1) + c$.

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