How do you integrate int x(5^(-x^2))dx?

Nov 28, 2016

The answer is $= - \left(\frac{1}{2}\right) {5}^{- {x}^{2}} / \ln 5 + C$

Explanation:

We use the substitution

$u = - {x}^{2}$

$\mathrm{du} = - 2 x \mathrm{dx}$

$x \mathrm{dx} = \frac{- \mathrm{du}}{2}$

Therefore,

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

Let, $y = {5}^{u}$

Then taking the logarithm

$\ln y = u \ln 5$

$y = {e}^{u \ln 5}$

$\int {5}^{u} \mathrm{du} = \int {e}^{u \ln 5} \mathrm{du} = {e}^{u \ln 5} / \ln 5 = \frac{y}{\ln} 5 = {5}^{u} / \ln 5$

Therefore,

$\int x \left({5}^{- {x}^{2}}\right) \mathrm{dx} = - \left(\frac{1}{2}\right) {5}^{- {x}^{2}} / \ln 5 + C$