MCQ
$\int_{}^{} {{x^3}{e^{{x^2}}}dx = } $
  • A
    $\frac{1}{2}({x^2} + 1){e^{{x^2}}} + c$
  • B
    $({x^2} + 1){e^{{x^2}}} + c$
  • $\frac{1}{2}({x^2} - 1){e^{{x^2}}} + c$
  • D
    $({x^2} - 1){e^{{x^2}}} + c$

Answer

Correct option: C.
$\frac{1}{2}({x^2} - 1){e^{{x^2}}} + c$
c
(c) Put ${x^2} = t \Rightarrow 2x\,dx = dt,$ then $\int_{}^{} {{x^3}{e^{{x^2}}}dx} = \frac{1}{2}\int_{}^{} {t{e^t}dt} $
$ = \frac{1}{2}\left[ {t{e^t} - {e^t}} \right] + c$$ = \frac{1}{2}{e^{{x^2}}}({x^2} - 1) + c.$

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