int x^3 e^{x^2} dx = | vedclass

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(C) Let $I = \int x^3 e^{x^2} dx$. Substitute $t = x^2$,then $dt = 2x dx$,which implies $x dx = \frac{1}{2} dt$. The integral becomes: $I = \int x^2 e

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int x^3 e^{x^2} dx$. Substitute $t = x^2$,then $dt = 2x dx$,which implies $x dx = \frac{1}{2} dt$. The integral becomes: $I = \int x^2 e^{x^2} (x dx) = \int t e^t (\frac{1}{2} dt) = \frac{1}{2} \int t e^t dt$. Using integration by parts,$\int u dv = uv - \int v du$,where $u = t$ and $dv = e^t dt$: $I = \frac{1}{2} [t e^t - \int e^t dt] = \frac{1}{2} [t e^t - e^t] + c$. Substituting back $t = x^2$: $I = \frac{1}{2} e^{x^2} (x^2 - 1) + c$.

$\int x^3 e^{x^2} dx = $

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