Integration by parts, What I should do next

Question

I'm integrating a function by parts but I get stuck in a part.

Since here, What should I do next?

Answer 1

Note that $\frac{x^2}{x+1}=x+\frac{1}{x+1}-1$, then the integral shouldn't be hard at all.

Answer 2

Hint: $${x^2 \over x+1} = {(x+1)^2 \over x+1} - {2x + 1 \over x+1} = {(x+1)^2 \over x+1} - {2(x + 1) \over x+1} + {1\over x+1}$$

Answer 3

Notice that $$\frac{x^2}{x+1} =\frac{(x + 1 -1)^2}{x +1} = \frac{(x+1)^2 - 2(x+1) + 1}{x+1}$$

Answer 4

alright so u got the integral $\int \:x\ln \left(x+1\right)dx$ ?

let

$u=x+1\quad \quad du=1dx,\:\quad \:dx=1du$

sub dx for du, and x+1 for u

$=\int \:x\ln \left(u\right)du$

$u=x+1\quad \Rightarrow \quad \:x=u-1$

$=\int \left(u-1\right)\ln \left(u\right)du$

integration by parts:

$u=\ln \left(u\right),\:\:u'=\frac{1}{u},\:\:v'=\left(u-1\right),\:\:v=\frac{\left(u-2\right)u}{2}$

$=\ln \left(u\right)\frac{\left(u-2\right)u}{2}-\int \frac{1}{u}\frac{\left(u-2\right)u}{2}du =\frac{\left(u-2\right)u\ln \left(u\right)}{2}-\int \frac{u-2}{2}du =\frac{\left(u-2\right)u\ln \left(u\right)}{2}-\frac{\frac{u^2}{2}-2u}{2}$

sub back u=x+1 and add the integration constant and ur done.