Evaluate $\int_0^\infty \frac{dx}{x^2+2ax+b}$ For $a^2<b$, is there an identity of evaluating the following integral?

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

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.
 A: 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?
A: For the first one:
$$(x+a)^2=x^2+2ax+a^2$$
So
$$(x+a)^2+b-a^2=x^2+2ax+b$$
$$=(b-a^2)\left(\frac{(x+a)^2}{b-a^2}+1 \right)$$
$$=(b-a^2) \left( \left( \frac{x+a}{\sqrt{b-a^2}}   \right)^2+1 \right)$$
For $b-a^2 \neq 0$. In which case, for evaluating the integral, we enforce the substitution  $\tan {u}=\frac{x+a}{\sqrt{b-a^2}}$ and proceed from there. Or the substitution $t=\frac{x+a}{\sqrt{b-a^2}}$, $\sqrt{b-a^2} dt=dx$:
$$\int_{0}^{\infty} \frac{1}{(b-a^2) \left( \left( \frac{x+a}{\sqrt{b-a^2}}   \right)^2+1 \right)} dx$$
$$=\frac{\sqrt{b-a^2}}{b-a^2} \int_{\frac{a}{\sqrt{b-a^2}}}^{\infty} \frac{1}{t^2+1} dt$$
$$=\frac{\sqrt{b-a^2}}{b-a^2}\left(\frac{\pi}{2}-\arctan (\frac{a}{\sqrt{b-a^2}}) \right)$$
$$=\frac{\text{arccot} (\frac{a}{\sqrt{b-a^2}})}{\sqrt{b-a^2}}$$
The second integral can be found by the method already mentioned, i.e. finding:
$$\frac{\partial}{\partial b} \left(-\frac{\text{arccot} (\frac{a}{\sqrt{b-a^2}})}{\sqrt{b-a^2}} \right)$$
In fact if we take:
$$\begin{equation*} I_n = \int^{\infty}_{0}\frac{1}{(x^2 + 2ax + b)^n} \mathrm{d}x\end{equation*}$$
Like zain did,
Then for $n \geq 2$:
$$I_n=\frac{\partial^{n-1}}{\partial b} \left(\frac{(-1)^{n-1}}{(n-1)!}\frac{\text{arccot} (\frac{a}{\sqrt{b-a^2}})}{\sqrt{b-a^2}} \right)$$
Because:
$$\frac{\partial^{n}}{\partial x} \frac{1}{x+c}=\frac{(-1)^nn!}{(x+c)^{n+1}}$$
Thus:
$$I_n(a,b)=\frac{(-1)^{n-1}}{(n-1)!}\frac{\partial^{n-1}}{\partial b_0} \left(\frac{\text{arccot} (\frac{a_0}{\sqrt{b_0-a_0^2}})}{\sqrt{b_0-a_0^2}} \right) \biggr \rvert_{(a,b)}$$
