How do you integrate $\int x^{2} e^{x^{3}}$ $int x^2e^(x^3)$ by parts?

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Answer 1 · 2016-12-18T01:50:18

Using integration by parts is very artificial for this integral. Substitution is much more reasonable.

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

Let $u = x^{3}$ $u = x^3$ . This makes $d u = 3 x^{2} d x$ $du = 3x^2 dx$ .

The integral becomes

$\frac{1}{3} \int e^{x^{3}} (3 x^{2} d x) = \frac{1}{3} \int e^{u} d u$ $1/3 int e^(x^3) (3x^2dx) = 1/3 int e^u du$

$= \frac{1}{3} e^{u} + C$ $= 1/3 e^u + C$

$= \frac{1}{3} e^{x^{3}} + C$ $= 1/3 e^(x^3) + C$

If I am told that I must use parts ,

I'll let $u = 1$ $u = 1$ and $d v = x^{2} e^{x^{3}} d x$ $dv = x^2e^(x^3) dx$

so that $d u = 0 d x$ $du = 0 dx$ and $v = \frac{1}{3} e^{x^{3}}$ $v = 1/3e^(x^3)$ .

And

$u v = \int v d u = 1 \cdot \frac{1}{3} e^{x^{3}} - \int \frac{1}{3} e^{x^{3}} \cdot 0 d u$ $uv=int v du = 1 * 1/3e^(x^3) - int 1/3e^(x^3) * 0 du$

$= \frac{1}{3} e^{x^{3}} + C$ $= 1/3 e^(x^3) + C$ .

How do you integrate ∫x2ex3int x^2e^(x^3) by parts?