Question

Is this correct?
\[\int \frac{x^2 dx}{( 1 + x)^2} \]Let u = 1+x, du = dx , \[x^2 = (u-1)^2\]\[\int \frac{x^2 dx}{( 1 + x)^2}\]\[ = \int \frac{(u-1)^2 du}{( u)^2} \]\[ = \int \frac{(u^2-2u+1) du}{( u)^2} \]\[ = \int 1 - \frac{2}{(u)} + \frac{1}{( u)^2} du\]\[ = u - 2ln|u| -u^{-1} +c = (x+1) - 2ln|x+1| - \frac{1}{x+1}+C\]

Answer 1

looks good to me... but hey, it's just me... :)

Answer 2

yep... nice substitution...

Answer 3

I used substitution only because I don't know partial fraction :(
Thanks all !

Answer 4

thankyou you all are life savers