Question

Integrate by parts:

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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)

Answer 5

You'll need an extra 2 in there.

Answer 6

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

Answer 7

@nexis . r u still there?

Answer 8

yes im trying it out right now

Answer 9

thanks your answer completely cleared this up for me @helder_edwin

Answer 10

i was confused about what exactly to make dv

Answer 11

u r welcome