Intergate $\int \frac{x}{(x^2-3x+17)^2}\ dx$ 
$$\int \frac{x}{(x^2-3x+17)^2}\ dx$$

My attempt:
$$\int \frac{x}{(x^2-3x+17)^2}\ dx=\int \frac{x}{\left((x-\frac{3}{2})^2+\frac{59}{4}\right)^2}\ dx$$
let $u=x-\frac{3}{2}$
$du=dx$
$$\int \frac{u+\frac{3}{2}}{\left((u)^2+\frac{59}{4}\right)^2}\ du$$
How can I continue from here?

Separate the integral into two integrals: 
$\displaystyle\int\frac{u}{u^2 + \left(\sqrt{\frac{59}{4}}\right)^{2}} + \displaystyle\int\frac{\frac{3}{2}}{u^2 + \left(\sqrt{\frac{59}{4}}\right)^{2}}$ 
Use another $u$-substitution on the first integral and then use trig substitution on the second one. 
