For which values of $\displaystyle p$ the integral $\displaystyle \int^2_1 \frac{x}{(x-1)^p}dx$ converges?
Thank you...
$\displaystyle \int^2_1\frac{x}{(x-1)^p}dx<\int^2_1\frac{2}{(x-1)^p}dx$
Now, $\displaystyle \int\frac{2}{(x-1)^p}dx=\frac{1}{1-p}\frac{1}{(x-1)^{p-1}}$
By some theorem...
If $\displaystyle p-1<1$ or $\displaystyle p<2$ then the above integral converges.
...with substetution $\displaystyle x-1=u$.
$\displaystyle \int\frac{x}{(x-1)^p}dx=\int\frac{u+1}{(u)^p}du=\int\frac{1}{u^{p-1}}du+\int\frac{1}{u^p}du=\frac{1}{u^{p-2}}\frac{1}{2-p}+\frac{1}{u^{p-1}}\frac{1}{1-p}$
...then:
$\displaystyle p-2<1 $ and $\displaystyle p-1<1$
or:
$\displaystyle p<3 $ and $\displaystyle p<2$
Therefore:
$\displaystyle p<2$