Solving Integral Problem: \int\frac{x*e^x}{(x+1)^2}dx

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The problem is given as
\int\frac{x*e^x}{(x+1)^2}dx

I did u substitution with u=(x+1) and du=dx

which gives me \int\frac{(u-1)*e^{u-1}}{u^2}

simplifies to \int\frac{u*e^{u-1}-e^{u-1}}{u^2}

Then I separated it into two integrals

\int\frac{e^{u-1}}{u}-\int\frac{e^{u-1}}{u^2}


Now I'm stuck. I tried doing these separate integrals by parts, but it doesn't seem to be working for me. Am I going in the complete wrong direction with this? Any help would be appreciated.
 
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Everything looks good to me so far, what do you get when you integrate the second term by parts once?
 
Oh wow. I got caught up on the fact that I couldn't do the first term and didn't realize that I don't have to! The integral of the second term takes care of that for me. Thank you.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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