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

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(B) Let $I = \int x^{3} e^{x^{2}} dx$. Substitute $x^{2} = t$,then $2x dx = dt$,which implies $x dx = \frac{1}{2} dt$. Substituting these into the int

Vedclass · Accepted Answer

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

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(B) Let $I = \int x^{3} e^{x^{2}} dx$. Substitute $x^{2} = t$,then $2x dx = dt$,which implies $x dx = \frac{1}{2} dt$. Substituting these into the integral,we get $I = \frac{1}{2} \int t e^{t} dt$. Using integration by parts,$\int u dv = uv - \int v du$,where $u = t$ and $dv = e^{t} dt$. Then $du = dt$ and $v = e^{t}$. So,$I = \frac{1}{2} [t e^{t} - \int e^{t} dt] = \frac{1}{2} [t e^{t} - e^{t}] + c$. Factoring out $e^{t}$,we get $I = \frac{1}{2} e^{t}(t - 1) + c$. Substituting back $t = x^{2}$,we obtain $I = \frac{1}{2} e^{x^{2}}(x^{2} - 1) + c$.

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

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