# How do you integrate int y/(e^(2y)) by integration by parts method?

Dec 9, 2016

$I = c - \frac{y {e}^{- 2 y}}{2} - {e}^{- 2 y} / 4$

#### Explanation:

We have

$I = \int \frac{y}{e} ^ \left(2 y\right) \mathrm{dy}$

With some simple rewriting, based on the properties of exponentials $\frac{1}{a} ^ \left(x\right) = {a}^{- x}$

$I = \int y {e}^{- 2 y} \mathrm{dy}$

If we say

$u = y$ so $\mathrm{du} = \mathrm{dy}$
$\mathrm{dv} = {e}^{- 2 y} \mathrm{dy}$ so $v = - {e}^{- 2 y} / 2$

By the integration by parts formula

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

$I = - \frac{y {e}^{- 2 y}}{2} - \int - {e}^{- 2 y} / 2 \mathrm{dy}$

Putting the constants outside of the integral

$I = - \frac{y {e}^{- 2 y}}{2} + \frac{1}{2} \int {e}^{- 2 y} \mathrm{dy}$
$I = c - \frac{y {e}^{- 2 y}}{2} - {e}^{- 2 y} / 4$