Can I use integration by parts recursively on this?

[cscott]

Can I use integration by parts recursively on this?

\int (xe^x)(x+1)^{-2}

 

[matt grime]

Have you tried?

 

[cscott]

Yeah

... = \frac{-xe^x}{x + 1} - \int \frac{e^x}{x + 1} \cdot dx

Right so far?

 

[arildno]

Nope.
\frac{d}{dx}xe^{x}=e^{x}(x+1)
There is at least one other flaw with your work.

 

[cscott]

Oops... don't know what I was thinking there. I got it now, thanks.

 

[Orion1]

Can I use integration by parts recursively on this?

Affirmative.

Formula for integration by parts by Substitution Rule:
\int u dv = uv - \int v du
u = xe^x \; \; \; dv = \frac{1}{(x + 1)^2} dx
du = e^x (x + 1) dx \; \; \; v = \left( - \frac{1}{x + 1} \right)
\int (xe^x) \left[\frac{1}{(x + 1)^2} \right] dx = (xe^x) \left(-\frac{1}{x + 1} \right) - \int \left(-\frac{1}{x + 1} \right)[e^x(x + 1)] dx
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[jacy]

Can this

\int (xe^x) \left[\frac{1}{(x + 1)^2} \right] dx = (xe^x) \left(-\frac{1}{x + 1} \right) - \int \left(-\frac{1}{x + 1} \right)[e^x(x + 1)] dx

be further simplified to
\int (xe^x) \left[\frac{1}{(x + 1)^2} \right] dx = \frac {e^x}{x+1}

 

[Orion1]

Affirmative
\left(xe^x\right)\left(-\frac{1}{x + 1}\right) - \int \left(-\frac{1}{x + 1}\right)\left[e^x\left(x + 1\right)\right] dx = - \frac{xe^x}{x + 1} + \int e^x dx

- \frac{xe^x}{x + 1} + \int e^x dx = - \frac{xe^x}{x + 1} + e^x = e^x \left(- \frac{x}{x + 1} + 1\right) = \frac{e^x}{x + 1}

\boxed{\int \left(xe^x\right)\left[\frac{1}{\left(x + 1\right)^2}\right] dx = \frac{e^x}{x + 1}}
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