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

• Nov 19th 2008, 02:47 AM
Craka
integral of (x+3)e^-x by parts
Question is to integrate the following integral by parts

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

Is this right?

$
\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}
$

The answer is given as

$
- e^{ - x} (x + 4) + K
$

but I can't see how
• Nov 19th 2008, 03:11 AM
Math_Helper
• Nov 19th 2008, 03:14 AM
Math_Helper
• Nov 19th 2008, 03:16 AM
Craka
Sorry, was I right in my answer?
• Nov 19th 2008, 03:19 AM
Math_Helper
all its right, but you didn't finish your answer
• Nov 19th 2008, 03:48 AM
Craka
how do you mean I didn't finish my answer?
• Nov 19th 2008, 03:53 AM
mr fantastic
Quote:

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

$= - (x + 3)e^{ - x} - e^{ - x} + K = -x e^{-x} - 3e^{-x} - e^{-x} + K = -xe^{-x} - 4 e^{-x} + K$ $= -e^{-x} (x + 4) + K$.
• Nov 19th 2008, 01:21 PM
Craka
whoops. Thanks.