$$ \int_{0}^{\infty} \frac{1+t^2}{(1+a^2)t^4+2t^2+1} dt $$ where $a \in \mathbb{R}$
I already divided by $t^2$ on both the numerator and denominator, and used substitution $x = t - \frac{1}{t}$. It didn't work because of the coefficient $1+a^2$ in $t^2$.
After dividing by $t^2$, take $(1+a^2)$ outside the denominator. We get:
$$ I = \frac{1}{a^2+1} \cdot \int_0^{\infty} \frac{1+\frac{1}{t^2}}{t^2+\frac{2}{1+a^2} + \frac{1}{(1+a^2)t^2}} dt$$
Let
$$\beta = \frac{1}{1+a^2}$$
Rewriting, we get
$$ I = \beta \cdot \int_0^{\infty} \frac{1+\frac{1}{t^2}}{t^2+ 2\beta + \frac{\beta}{t^2}} dt$$
Now let
$$ 1 + \frac{1}{t^2} = \lambda_1 \cdot ( 1 + \frac{\sqrt{\beta}}{t^2}) + \lambda_2 \cdot ( 1 - \frac{\sqrt{\beta}}{t^2})$$
Now, all you have to do is find the values of $\lambda_1$ and $\lambda_2$, split the integral into two separate integrals - the first one, you substitute $ x = t - \frac{\sqrt{\beta}}{t}$ and for the second one you can substitute $ x = t + \frac{\sqrt{\beta}}{t}$ and proceed.
It is easy to compute the integral after this substitution.