Integrate by parts:

Question

Integrate by parts:

anonymous · Answer

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

hartnn · Answer

the first thing i can think of is to put 1+x^2 as u
have u tried it?

tkhunny · Answer

Really big hint: $$\frac{d}{dx}\frac{1}{1+x^{2}} = ??$$

anonymous · Answer

hmm let me reformulate my question maybe because I also have a more general one and there was something else I was confused about, though this may be a valid approach. In one solution approach I saw this integral being broke down into $$x^2*\frac{x}{(1+x^2)^2}dx$$, then u=x^2 and v'=1+x^2. In this case, the xdx and the squared term over the denominator are not substituted.
My question is: can you still do integration by parts in such a case? (Thanks for your input though, I'll try that right now)

tkhunny · Answer

You'll need an extra 2 in there.

helder_edwin · Answer

what u proposed looks fine.
$$ \large u=x^2\qquad du=2x\,dx $$
$$ \large dv=\frac{x}{(1+x^2)^2}\qquad v=-\frac{1}{2}\frac{1}{1+x^2} $$

helder_edwin · Answer

@nexis . r u still there?

anonymous · Answer

yes im trying it out right now

anonymous · Answer

thanks your answer completely cleared this up for me @helder_edwin

anonymous · Answer

i was confused about what exactly to make dv

helder_edwin · Answer

u r welcome