Hi there,

I am not too sure how to go about the question 'integrate xlnx dx'
the answer is meant to be (x^2lnx)/2-x^2/4+C

Use integration by parts. Let:

$u=\ln(x)\,\therefore\.du=\frac{1}{x}\,dx$

$dv=x\,dx\,\therefore\,v=\frac{x^2}{2}$

and then:

$\int x\ln(x)\,dx=uv-\int v\,du$

Oh, ok!! I think I had messed up the 'u', 'du', 'dv' and 'dv' before.

Thank you so much for your help

when integrating -x^2(1/x dx), why does 2 become a numerator (e.g. (2-x^2)/4)

I'm not sure what you are asking, but:

$-\int v\,du=-\int \frac{x^2}{2}\cdot\frac{1}{x}\,dx=-\frac{1}{2}\int x\,dx=-\frac{x^2}{4}+C$