_ ^ x^5 . e^ x^2 dx = - Vidyadip

MCQ

$\int_{}^{} {{x^5}.{e^{{x^2}}}dx = } $

✓
$\frac{1}{2}{x^4}{e^{{x^2}}} - {x^2}{e^{{x^2}}} + {e^{{x^2}}} + c$
B
$\frac{1}{2}{x^4}{e^{{x^2}}} + {x^2}{e^{{x^2}}} + {e^{{x^2}}} + c$
C
$\frac{1}{2}{x^4}{e^{{x^2}}} - {x^2}{e^{{x^2}}} - {e^{{x^2}}} + c$
D
None of these

✓

Answer

Correct option: A.

$\frac{1}{2}{x^4}{e^{{x^2}}} - {x^2}{e^{{x^2}}} + {e^{{x^2}}} + c$

a
(a) Put ${x^2} = t \Rightarrow 2x\,dx = dt,$ then
$\int_{}^{} {{x^5}{e^{{x^2}}}dx} = \frac{1}{2}\int_{}^{} {{t^2}{e^t}dt} = \frac{1}{2}\left[ {{e^t}{t^2} - 2\int_{}^{} {t{e^t}dt} } \right] + c$
$ = \frac{{{t^2}{e^t}}}{2} - \left[ {t{e^t} - {e^t}} \right] + c = \frac{1}{2}{x^4}{e^{{x^2}}} - {x^2}{e^{{x^2}}} + {e^{{x^2}}} + c.$

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