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

MCQ

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

✓
$\left( {\frac{{x - 1}}{{x + 1}}} \right){e^x} + c$
B
${e^x}\left( {\frac{{x + 1}}{{x - 1}}} \right) + c$
C
${e^x}(x + 1)(x - 1) + c$
D
None of these

✓

Answer

Correct option: A.

$\left( {\frac{{x - 1}}{{x + 1}}} \right){e^x} + c$

a
(a)$\int_{}^{} {\frac{{{e^x}({x^2} + 1)}}{{{{(x + 1)}^2}}}\,dx} = \int_{}^{} {\frac{{{e^x}({x^2} - 1 + 2)}}{{{{(x + 1)}^2}}}\,dx} $
$ = \int_{}^{} {{e^x}\left[ {\frac{{x - 1}}{{x + 1}} + \frac{2}{{{{(x + 1)}^2}}}} \right]} \,dx = \int_{}^{} {{e^x}[f(x) + f'(x)]\,dx} $
where $f(x) = \frac{{x - 1}}{{x + 1}}$ and $f'(x) = \frac{2}{{{{(x + 1)}^2}}} = {e^x}\left( {\frac{{x - 1}}{{x + 1}}} \right) + c$.

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