# How do you integrate int x^c e^(x^d)dx, where c>d, using integration by parts?

Oct 4, 2016

Use the recurrence formula

${I}_{d + k} + \frac{k + 1}{\mathrm{dI}} _ k = \frac{1}{\mathrm{dx}} ^ \left(k + 1\right) {e}^{{x}^{d}}$

with initial $k = c - d$ and

${I}_{d - 1} = \frac{1}{\mathrm{de}} ^ \left({x}^{d}\right)$

#### Explanation:

We use $n , m$ instead of $c , d$,
.

$\frac{d}{\mathrm{dx}} \left({x}^{n} {e}^{{x}^{m}}\right) = n {x}^{n - 1} {e}^{m} + m {x}^{n + m - 1} {e}^{{x}^{m}}$

Calling now

${I}_{n} = \int {x}^{n} {e}^{{x}^{m}} \mathrm{dx}$ we have the recurrence relationship

$m {I}_{n + m - 1} + n {I}_{n - 1} = {x}^{n} {e}^{{x}^{m}}$

with

${I}_{m - 1} = \int {x}^{m - 1} {e}^{{x}^{m}} \mathrm{dx} = \frac{1}{m} {e}^{{x}^{m}}$ or calling $k = n - 1$

$m {I}_{m + k} + \left(k + 1\right) {I}_{k} = {x}^{k + 1} {e}^{{x}^{m}}$

Finally

${I}_{m + k} + \frac{k + 1}{m} {I}_{k} = \frac{1}{m} {x}^{k + 1} {e}^{{x}^{m}}$

Example. $n = 5 , m = 3$ so $k = 2$ and we want

${I}_{m + 2} = {I}_{5} = \int {x}^{5} {e}^{{x}^{3}} \mathrm{dx}$

we know that ${I}_{m - 1} = {I}_{2}$ and

${I}_{2} = \frac{1}{3} {e}^{{x}^{3}}$ so

${I}_{m + 2} + \frac{2 + 1}{m} {I}_{2} = \frac{1}{m} {x}^{3} {e}^{{x}^{3}}$ so

${I}_{5} + \frac{3}{3} {I}_{2} = \frac{1}{3} {x}^{3} {e}^{{x}^{3}}$ or

${I}_{5} = \frac{1}{3} {x}^{3} {e}^{{x}^{3}} - \frac{1}{3} {e}^{{x}^{3}} = \frac{1}{3} \left({x}^{3} - 1\right) {e}^{{x}^{3}}$