calculus - Evaluate $int_0^infty frac{dx}{x^2+2ax+b}$

For $a^2

$$\int_0^\infty \frac{dx}{x^2+2ax+b}$$

What about:

$$\int_0^\infty \frac{dx}{(x^2+2ax+b)^2}$$

My attempt is using partial fractions and completing the square, but I still failed to obtain a nice result.

Answer

Let the first integral be $I$ and the second one be $J$, then by putting $x=y-a$ and $y=z\sqrt{b-a^2}$ we have
\begin{align}
I(a,b)&=\int_0^{\infty}\frac{dx}{(x+a)^2+b-a^2}\\[10pt]
&= \int_a^{\infty}\frac{dy}{y^2+b-a^2}\\[10pt]
&=\frac{1}{\sqrt{b-a^2}}\int_{\large\frac{a}{\sqrt{b-a^2}}}^{\infty}\frac{dz}{z^2+1}\\[10pt]
&=\frac{1}{\sqrt{b-a^2}}\left(\frac{\pi}{2}-\arctan\left(\frac{a}{\sqrt{b-a^2}}\right)\!\right)\\[15pt]
\end{align}

and the 2nd integral is just a first derivative of $I$ with respect to $b$ times $-1$
\begin{equation}
\\[15pt]J(a,b)=-\frac{\partial I}{\partial b}=\frac{\partial }{\partial b}\left[\frac{1}{\sqrt{b-a^2}}\left(\arctan\left(\frac{a}{\sqrt{b-a^2}}\right)-\frac{\pi}{2}\right)\right]\\[10pt]
\end{equation}
Can you take it from here?