Find $\lim\limits_{t \to \infty} \int_{0}^{t} \frac{\mathrm dx}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}$ 
Find $$\lim_{t \to \infty} \int_{0}^{t} \frac{1}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}\mathrm dx$$ where $a,b,c$ are strict positive and dinstinct real numbers.

I know it should be something with arctangent but I don't know how to get there. Can somebody give me some tips, please?

Using partial fraction decomposition, we have that
$$\begin{align*}
\frac1{(x^2+a^2)(x^2+b^2)(x^2+c^2)}&=\frac1{c^2-a^2}\left(\frac1{(x^2+a^2)(x^2+b^2)}-\frac1{(x^2+b^2)(x^2+c^2)}\right)\\&=\frac1{c^2-a^2}\left(\frac{1}{b^2-a^2}\left(\frac1{x^2+a^2}-\frac1{x^2+b^2}\right)\right)\\&-\frac1{c^2-a^2}\left(\frac{1}{c^2-b^2}\left(\frac1{x^2+b^2}-\frac1{x^2+c^2}\right)\right).
\end{align*}$$ 
Using $\lim_{t\to\infty}\int_0^t \frac{1}{x^2+\alpha^2}\mathrm dx=\frac{\pi}{2\alpha}$, $\ \alpha>0$, we can obtain the integral:
$$\begin{align*}
I&=\frac1{c^2-a^2}\frac1{b^2-a^2}\left(\frac1{a}-\frac1b\right)\frac{\pi}2-\frac1{c^2-a^2}\frac1{c^2-b^2}\left(\frac1{b}-\frac1c\right)\frac{\pi}2
\\&=\frac1{c^2-a^2}\frac{\pi}{2ab(a+b)}-\frac1{c^2-a^2}\frac{\pi}{2bc(b+c)}\\
&=\frac{\pi(a+b+c)}{2abc(a+b)(b+c)(c+a)}.
\end{align*}$$
