How do you integrate $\int x^{3} e^{x^{2} - 1} d x$ $int x^3e^(x^2-1)dx$ using integration by parts?

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Answer 1 · 2015-11-09T15:51:18

See the explanation.

$\int x^{3} e^{x^{2} - 1} d x$ $int x^3e^(x^2-1)dx$

Integrateing $x^{3}$ $x^3$ and differentiating $e^{x^{2} - 1}$ $e^(x^2-1)$ would give us MORE $x$ $x$ 's, so let's try the other way around first.

To integrate $e^{x^{2} - 1}$ $e^(x^2-1)$ , we'll need another $x$ $x$ so that we can substitute.

So,
$\int x^{3} e^{x^{2} - 1} d x = \int x^{2} [x e^{x^{2} - 1} d x]$ $int x^3e^(x^2-1)dx = int x^2[xe^(x^2-1)dx]$

Let $u = x^{2}$ $u = x^2$ and $d v = x e^{x^{2} - 1} d x$ $dv = xe^(x^2-1)dx$ .

We get $d u = 2 x d x$ $du = 2x dx$ and $v = \frac{1}{2} e^{x^{2} - 1}$ $v = 1/2e^(x^2-1)$ .

Our integral becomes

$= \frac{1}{2} x^{2} e^{x^{2} - 1} - \int x e^{x^{2} - 1} d x$ $= 1/2x^2e^(x^2-1) - intxe^(x^2-1)dx$ .

The integral may again be evaluated by substitution.

$= \frac{1}{2} x^{2} e^{x^{2} - 1} - \frac{1}{2} e^{x^{2} - 1} + C$ $= 1/2x^2e^(x^2-1) - 1/2e^(x^2-1) +C$ .

How do you integrate ∫x3ex2−1dxint x^3e^(x^2-1)dx using integration by parts?