# How do you find the indefinite integral of int x(5^(-x^2))?

##### 1 Answer
Dec 12, 2016

The answer is $= - {5}^{- {x}^{2}} / \left(2 \ln 5\right) + C$

#### Explanation:

We do the integral by substitution

Let $u = {x}^{2}$, then, $\mathrm{du} = 2 x \mathrm{dx}$

$\int x \left({5}^{- {x}^{2}}\right) \mathrm{dx} = \frac{1}{2} \int \left({5}^{- u}\right) \mathrm{du}$

Let $v = {5}^{- u}$

$\ln v = \ln {5}^{- u} = - u \ln 5$

$v = {e}^{- u \ln 5}$

Therefore,

$\int x \left({5}^{- {x}^{2}}\right) \mathrm{dx} = \frac{1}{2} \int \left({5}^{- u}\right) \mathrm{du} = \frac{1}{2} \int {e}^{- u \ln 5} \mathrm{du}$

$= - {e}^{- u \ln 5} / \left(2 \ln 5\right) = - {5}^{- u} / \left(2 \ln 5\right) = - {5}^{- {x}^{2}} / \left(2 \ln 5\right) + C$