<s>[INST] How to solve the given integral of the type: $\int_{0}^{\infty}\tfrac{1}{(a + b x)(c + d x) (1 + p x)}dx$ Can anyone give a hint how to solve integral of the following type: 
$$\int_{0}^{\infty}\dfrac{1}{(a + b x)(c + d x) (1 + p x)}dx$$
The problem is if we proceed with partial fraction, we get
$$\dfrac{1}{(a + bx)(c + dx)(1 + px)} = \dfrac{b^2}{(bc - ad)(b - ap)(a + bx)} + \dfrac{d^2}{(bc - ad)(cp - d)(c + dx)}- \dfrac{p^2}{(b - ap)(cp-d)(1 + px)}$$
Now solving the integral for the first part, 
$$\int_{0}^{\infty}\dfrac{b^2}{(bc - ad)(b - ap)(a + bx)} dx = \left.\dfrac{b \ln(a + bx)}{(bc - ad)(b - ap)}\right\vert_0^{\infty}$$
Hence a $\ln(\infty)$ term is encountered. Same is the case with all the three parts. 
Any hints appreciated. 
Edit: The final solution using Mathematica:

 [/INST] For simplicity, write this as $$J = \dfrac{1}{bdp} \int_0^\infty \dfrac{dx}{(A+x)(B+x)(C+x)}$$
where $A = a/b$, and $B = c/d$, $C = 1/p$.  We're assuming $a,b,c,d,p > 0$ and $A,B,C$ are all distinct.  The partial fraction decomposition of the integrand is
$$ F(x) = \frac{1}{(A-B)(A-C)(A+x)} + \frac{1}{(B-A)(B-C)(B+x)} + \frac{1}{(C-A)(C-B)(C+x)}$$
so that
$$ J = \frac{1}{bdp} \lim_{R \to \infty} \left(\frac{\ln(A+R)-\ln(A)}{(A-B)(A-C)} + \frac{\ln(B+R)-\ln(B)}{(B-A)(B-C)} + \frac{\ln(C+R)-\ln(C)}{(C-A)(C-
B)}\right) $$
Now write $\ln(A+R) = \ln(R) + \ln(1+A/R)$ etc, and notice that (as Achille Hui remarked) the coefficients of $\ln(R)$ cancel (as they must, because the improper integral is known to converge), while $\ln(1+A/R), \ln(1+B/R), \ln(1+C/R) \to 0$
as $R \to \infty$.
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