_ ^ e^ ^ - 1 x ( 1 + x + x^2 1 + x^2 ) ;dx is equal to - Vidyadip

MCQ

$\int_{}^{} {{e^{{{\tan }^{ - 1}}x}}} \left( {\frac{{1 + x + {x^2}}}{{1 + {x^2}}}} \right)\;dx$ is equal to

✓
$x{e^{{{\tan }^{ - 1}}x}} + c$
B
${x^2}{e^{{{\tan }^{ - 1}}x}} + c$
C
$\frac{1}{x}{e^{{{\tan }^{ - 1}}x}} + c$
D
None of these

✓

Answer

Correct option: A.

$x{e^{{{\tan }^{ - 1}}x}} + c$

a
(a) Putting ${\tan ^{ - 1}}x = t$ and $\frac{{dx}}{{1 + {x^2}}} = dt,$ we get
$\int_{}^{} {{e^{{{\tan }^{ - 1}}x}}\left( {\frac{{1 + x + {x^2}}}{{1 + {x^2}}}} \right)} \,dx = \int_{}^{} {{e^t}(\tan t + {{\sec }^2}t)\,dt} $
$ = {e^t}\tan t + c = x\,{e^{{{\tan }^{ - 1}}x}} + c$
$\left[ {{\rm{Using }}\int_{}^{} {{e^x}\left\{ {f(x) + f'(x)} \right\}dx = {e^x}f(x) + C} } \right]$.

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