Partial Fractions Integral

Quote:

Originally Posted by Chicken Enchilada

These partial fractions integrals have been pretty easy so far. It's really easy once you get the problem setup the correct way. I'm having trouble with this one though.

$\int \frac{3x + 2}{x^3 + 3x^2 + 3x +1}dx$
I tried factoring it but either way you end up with an extra addition on the end. Maybe there is a way to factor it so that the top reduces out?

What am I missing?

Thanks,
Justin

$ \frac{3x + 2}{x^3 + 3x^2 + 3x +1} = $

$ \frac{3x + 2}{(x+1)^3} = $

$ \frac{2(x+1) + x}{(x+1)^3} = $

$ \frac{2}{(x+1)^2} + \frac{x}{(x+1)^3} $

$\int \frac{2}{(x+1)^2} \, dx + \int \frac{x}{(x+1)^3} \, dx$

$\int \frac{2}{(x+1)^2} \, dx + \int \frac{u-1}{u^3} \, du$

take it from here?