How do you use integration by parts to find intxe^-x dx?

Jul 28, 2014

$= - {e}^{-} x \left(1 + x\right) + c$, where c is a constant

Explanation

Using Integration by Parts,

$\int \left(I\right) \left(I I\right) \mathrm{dx} = \left(I\right) \int \left(I I\right) \mathrm{dx} - \int \left(\left(I\right) ' \int \left(I I\right) \mathrm{dx}\right) \mathrm{dx}$

where $\left(I\right)$ and $\left(I I\right)$ are functions of $x$, and $\left(I\right)$ represents which will be differentiated and $\left(I I\right)$ will be integrated subsequently in the above formula

Similarly following for the problem,

$= x \cdot \int {e}^{-} x \mathrm{dx} - \int \left(\left(x\right) ' \int {e}^{-} x \mathrm{dx}\right) \mathrm{dx}$

$= x \cdot {e}^{-} \frac{x}{- 1} + \int {e}^{-} x \mathrm{dx}$

$= - x \cdot {e}^{-} x + {e}^{-} \frac{x}{- 1} + c$, where c is a constant

$= - x \cdot {e}^{-} x - {e}^{-} x + c$, where c is a constant

$= - {e}^{-} x \left(1 + x\right) + c$, where c is a constant