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$\int x^{3} \cdot e^{x^{2}} d x=$
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Verified Answer
The correct answer is:
$\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c$
Let $I=\int x^{3} e^{x^{2}} d x$
Put $x^{2}=t \Rightarrow 2 x d x=d t \Rightarrow x d x=\frac{1}{2} d t$
$\therefore \mathrm{I}=\frac{1}{2} \int \mathrm{te}^{\mathrm{t}} \mathrm{dt}$
$=\frac{1}{2}\left[t e^{t}-e^{t}\right]+c=\frac{1}{2} e^{t}(t-1)+c=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c$
Put $x^{2}=t \Rightarrow 2 x d x=d t \Rightarrow x d x=\frac{1}{2} d t$
$\therefore \mathrm{I}=\frac{1}{2} \int \mathrm{te}^{\mathrm{t}} \mathrm{dt}$
$=\frac{1}{2}\left[t e^{t}-e^{t}\right]+c=\frac{1}{2} e^{t}(t-1)+c=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c$
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