Find $a$ and $b$ such that the improper integral converges Find $a$ and $b$ so that the integral $$\int_1^\infty \frac{x}{(x-1)^a(1+x^2)^b}\, dx$$ converges. 

I know that $\int_1^\infty \frac{1}{x^p}$ converges if $p>1$ and I've tried integration by parts but neither has helped. Can someone give me a hint as to how I should approach this?

We have that
$$\int_1^\infty \frac{x}{(x-1)^a(1+x^2)^b}\, dx=\int_1^2 \frac{x}{(x-1)^a(1+x^2)^b}\, dx+\int_2^\infty \frac{x}{(x-1)^a(1+x^2)^b}\, dx$$
and
$$\int_1^2 \frac{x}{(x-1)^a(1+x^2)^b}\, dx=\int_0^1 \frac{y+1}{y^a(y^2+2y+2)^b}\, dy$$
which converges for $a<1$ since as $y \to 0^+$
$$ \frac{y+1}{y^a(y^2+2y+2)^b} \sim\frac1{y^{a}}$$
and
$$\int_2^\infty \frac{x}{(x-1)^a(1+x^2)^b}dx, \quad \frac{x}{(x-1)^a(1+x^2)^b}\sim\frac1{x^{a+2b-1}} \quad x\to \infty$$
converges for $a+2b-1>1 \iff b>1-\frac a 2$.
Therefore the given integral converges for $a<1$ and $b>1-\frac a 2$.
