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?
 A: 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:
$$\begin{align*}
I&=\frac1{c^2-a^2}\frac1{b^2-a^2}\left(\frac1{a}-\frac1b\right)\frac{\pi}2-\frac1{c^2-a^2}\frac1{c^2-b^2}\left(\frac1{b}-\frac1c\right)\frac{\pi}2
\\&=\frac1{c^2-a^2}\frac{\pi}{2ab(a+b)}-\frac1{c^2-a^2}\frac{\pi}{2bc(b+c)}\\
&=\frac{\pi(a+b+c)}{2abc(a+b)(b+c)(c+a)}.
\end{align*}$$
A: Hints
Method 1: Partial fractions Since $a, b, c$ are positive and distinct, a straightforward application of the method of partial fractions gives that the integrand is $$\sum \frac{1}{(a^2 - b^2)(a^2 - c^2)} \frac{1}{x^2 + a^2},$$ where the sum is over the cyclic permutations of the parameters $(a, b, c)$. By linearity we can write the integral as a linear combination of integrals of the form $$\int_0^{\infty} \frac{dx}{x^2 + k^2} ,$$ and like you say, these can be evaluated with the usual antiderivative formula $$\int \frac{dx}{x^2 + k^2} = \frac{1}{k} \arctan \frac{x}{k} + C .$$
Method 2: Contour integration Since the integrand in even, it coincides with $$\frac{1}{2} \int_{-\infty}^\infty \frac{dx}{(x^2 + a^2) (x^2 + b^2) (x^2 + c^2)}.$$ This suggests integrating over a semicircle $\Gamma_R$ of radius $R$ with diameter on the real axis and centered at zero, say, in the upper half-plane. For large $R$, the value of the integral over the circular arc is $O(R^{-5})$, so in the limit its contribution to the integral is zero and hence $$\frac{1}{2} \int_{-\infty}^\infty \frac{dx}{(x^2 + a^2) (x^2 + b^2) (x^2 + c^2)} = \lim_{R \to \infty} \frac{1}{2} \oint_{\Gamma_R} \frac{dz}{(z^2 + a^2) (z^2 + b^2) (z^2 + c^2)} ,$$ but (for large $R$, and since $a, b, c$ are positive and distinct) the poles inside $\Gamma_R$ are simple poles at $ia, ib, ic$, so we can apply the residue formula, which gives that the integral is $$\frac{1}{2} \cdot 2 \pi i (\operatorname{Res}(f, ia) + \operatorname{Res}(f, ib) + \operatorname{Res}(f, ic)),$$ where $f$ is the integrand.
A: I addressed this question in a general form a couple weeks back! Here it is:

Let $T$ be a finite set of non zero real values. Then we may define the definite integral $I(T)$ as:
\begin{equation}
    I(T) = \int_{-\infty}^{\infty} \prod_{t \in T} \frac{1}{x^2 + t^2}\:dx \nonumber 
\end{equation}
To address this integral we decompose the integrand using a partial fraction decomposition:
\begin{equation}
    \prod_{t \in T} \frac{1}{x^2 + t^2} = \sum_{t \in T} \frac{A_t}{x^2 + t^2}\nonumber 
\end{equation}
Where $A_t \in \mathbb{R}$. Now:
\begin{align}
    \prod_{t \in T} \frac{1}{x^2 + t^2} &= \sum_{t \in T} \frac{A_t}{x^2 + t^2} = \frac{A_{t_1}}{x^2 + t_1^2} + \frac{A_{t_2}}{x^2 + t_2^2} + \dots + \frac{A_{t_n}}{x^2 + t_n^2} \nonumber 
\end{align}
And so:
\begin{equation}
1 = \sum_{t \in T} A_t\prod_{k \in T \setminus \{t\}} x^2 + k^2 \nonumber
\end{equation}
Thus, 
\begin{equation}
    A_t = \left[\prod_{k \in T \setminus \{t\}} k^2 - t^2\right]^{-1} \nonumber
\end{equation}
We now can evaluate our integral $I(T)$:
\begin{align}
    I(T) &= \int_{-\infty}^{\infty} \prod_{t \in T} \frac{1}{x^2 + t^2}\:dx = \sum_{t \in T} A_t \int_{-\infty}^{\infty}\frac{1}{x^2 + t^2}\:dx   = \sum_{t \in T} A_t \left[\frac{1}{|t|}\arctan\left(\frac{x}{|t|}\right) \right]_{-\infty}^{\infty}  \nonumber \\
    &=\sum_{t \in T} A_t \cdot \frac{\pi}{|t|} = \pi \sum_{t \in T} \frac{1}{|t|} A_t \nonumber  =  \pi \sum_{t \in T} \left[|t| \prod_{k \in T \setminus \{t\}} k^2 - t^2\right]^{-1}
\end{align}
And So we arrive at:
\begin{equation}
    I(T) = \int_{-\infty}^{\infty} \prod_{t \in T} \frac{1}{x^2 + t^2}\:dx = \pi \sum_{t \in T} \frac{1}{|t|}\prod_{k \in T \setminus \{t\}}\frac{1}{k^2 - t^2} \nonumber 
\end{equation}

For your integral $T = \left\{a,b,c \right\}$
