# integral of (x+3)e^-x by parts

• Nov 19th 2008, 01:47 AM
Craka
integral of (x+3)e^-x by parts
Question is to integrate the following integral by parts

$\displaystyle \int {(x + 3)e^{ - x} dx}$

Is this right?

$\displaystyle \begin{array}{l} u = x + 3 \\ u' = 1 \\ v' = e^{ - x} \\ v = - e^{ - x} \\ \int {(x + 3)e^{ - x} dx} \\ = (x + 3) \times - e^{ - x} - \int {1 \times - e^{ - x} dx} \\ = - (x + 3)e^{ - x} - e^{ - x} +K \\ \end{array}$

$\displaystyle - e^{ - x} (x + 4) + K$
but I can't see how
• Nov 19th 2008, 02:11 AM
Math_Helper
• Nov 19th 2008, 02:14 AM
Math_Helper
• Nov 19th 2008, 02:16 AM
Craka
Sorry, was I right in my answer?
• Nov 19th 2008, 02:19 AM
Math_Helper
• Nov 19th 2008, 02:48 AM
Craka
how do you mean I didn't finish my answer?
• Nov 19th 2008, 02:53 AM
mr fantastic
Quote:

Originally Posted by Craka
how do you mean I didn't finish my answer?

$\displaystyle = - (x + 3)e^{ - x} - e^{ - x} + K = -x e^{-x} - 3e^{-x} - e^{-x} + K = -xe^{-x} - 4 e^{-x} + K$ $\displaystyle = -e^{-x} (x + 4) + K$.
• Nov 19th 2008, 12:21 PM
Craka
whoops. Thanks.