Trying to do the integral of xe^-8x using integration by parts.

if I Iet u=x, du=1 and v=-1/8(e^-8t), dv=e^-8x
and worked the answer out to -1/8(x*e^-8x)+1/64(e^-8x) but it's not correct.

Question

Trying to do the integral of xe^-8x using integration by parts.

if I Iet u=x, du=1 and v=-1/8(e^-8t), dv=e^-8x 
and worked the answer out to -1/8(x*e^-8x)+1/64(e^-8x) but it's not correct.

anonymous · Answer

$$I=\int\limits x e ^{-8x}dx=x \int\limits e ^{-8x}dx-\int\limits \frac{ d }{ dx }\left( x \right) \int\limits e ^{-8x}dx+c$$
$$=x \frac{ e ^{-8 x} }{ -8 }- \int\limits 1*\frac{ e ^{-8x} }{ -8 }dx+c$$
$$=-\frac{ 1 }{ 8 }x e ^{-8x}-\frac{ e ^{-8x} }{ -8*(-8) }+c$$
$$=-\frac{ 1 }{ 8 }x e ^{-8x}-\frac{ 1 }{ 64 }e ^{-8x}+c$$

theoreo · Answer

Hmmm ok, so my sign was wrong for the second part. Looks like I did the rest of it right though. Thanks Surjit

anonymous · Answer

yw