<s>[INST] 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.
 [/INST] 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:
$$I_n(a,b)=\frac{(-1)^{n-1}}{(n-1)!}\frac{\partial^{n-1}}{\partial b_0} \left(\frac{\text{arccot} (\frac{a_0}{\sqrt{b_0-a_0^2}})}{\sqrt{b_0-a_0^2}} \right) \biggr \rvert_{(a,b)}$$
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