The value of x (x + 1) ^2 dx is - Vidyadip

MCQ

The value of $\int {\frac{{\log x}}{{{{(x + 1)}^2}}}dx} $ is

✓
$\frac{{ - \log x}}{{x + 1}} + \log x - \log \,(x + 1)$
B
$\frac{{\log x}}{{\left( {x + 1} \right)}} + \log x - \log \,(x + 1)$
C
$\frac{{\log x}}{{x + 1}} - \log x - \log \,(x + 1)$
D
$\frac{{ - \log x}}{{x + 1}} - \log x - \log \,(x + 1)$

✓

Answer

Correct option: A.

$\frac{{ - \log x}}{{x + 1}} + \log x - \log \,(x + 1)$

a
(a)$\int {\frac{{\log x}}{{{{(x + 1)}^2}}}dx = \int {\log x\,{{(x + 1)}^{ - 2}}} } dx$
$ = \log x.\left\{ { - {{(x + 1)}^{ - 1}}} \right\}$$ - \int {\frac{1}{x}.\{ - {{(x + 1)}^{ - 1}}\} dx} $
$ = \frac{{ - \log x}}{{(x + 1)}} + \int {\frac{1}{{x(x + 1)}}dx} $$ = \frac{{ - \log x}}{{(x + 1)}} + \int {\left[ {\frac{1}{x} - \frac{1}{{x + 1}}} \right]dx} $
$ = \frac{{ - \log x}}{{x + 1}} + \log x - \log (x + 1)$.

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