_ ^ x - 1 (x + 1) ^2 ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{x - 1}}{{{{(x + 1)}^2}}}\;dx = } $

✓
$\log (x + 1) + \frac{2}{{x + 1}} + c$
B
$\log (x + 1) - \frac{2}{{x + 1}} + c$
C
$\frac{2}{{x + 1}} - \log (x + 1) + c$
D
None of these

✓

Answer

Correct option: A.

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

a
(a) $\int_{}^{} {\frac{{x - 1}}{{{{(x + 1)}^2}}}\,dx = \int_{}^{} {\frac{{x + 1 - 2}}{{{{(x + 1)}^2}}}} \,dx} $
$ = \int_{}^{} {\frac{1}{{x + 1}}\,dx} - \int_{}^{} {\frac{2}{{{{(x + 1)}^2}}}\,dx = \log (x + 1) + \frac{2}{{(x + 1)}} + c} $.

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