Problem in solving the improper integral. I am trying to solve the following improper integral $I$.
$$I = \int\limits_0^\infty\frac{1}{\left(x+1\right)\left(ax+1\right)\left(bx+1\right)} dx$$
However, I myself get the divergent answer. When I tried it through an online-calculator, it gives me something else. Can someone guide me through the following? I will really appreciate.
\begin{align*}
I
&=\lim_{x\to \infty}\frac{\left(a-1\right)b\ln\left(\left|bx+1\right|\right)+\left(a-ab\right)\ln\left(\left|ax+1\right|\right)+\left(b-a\right)\ln\left(\left|x+1\right|\right)}{\left(a-1\right)\left(b-1\right)\left(b-a\right)} \\
&\quad - \lim_{x\to 0}\dfrac{\left(a-1\right)b\ln\left(\left|bx+1\right|\right)+\left(a-ab\right)\ln\left(\left|ax+1\right|\right)+\left(b-a\right)\ln\left(\left|x+1\right|\right)}{\left(a-1\right)\left(b-1\right)\left(b-a\right)}.
\end{align*}
While solving the part for $\lim_{x\to0}$, I get $0$ as the output, but if I solve $\lim_{x\to\infty}$, I get a divergent term. While in actual if I solve it through an online calculator, I get the following answer.
$$I = \dfrac{\left(a-1\right)b\ln\left(b\right)-a\ln\left(a\right)b+a\ln\left(a\right)}{\left(a-1\right)\left(b-1\right)\left(b-a\right)}$$
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
I & = \int_{0}^{\infty}{\dd x \over
\pars{x + 1}\pars{ax + 1}\pars{bx + 1}}
\\[5mm] & =
\int_{0}^{\infty}\braces{2\int_{0}^{1}\dd u_{1}\int_{0}^{u_{1}}\dd u_{2}\,{1 \over \bracks{x + 1 + u_{1}\pars{a - 1}x + u_{2}\pars{b - a}x}^{\, 3}}}\dd x
\\[5mm] & =
2\int_{0}^{1}\dd u_{1}\int_{0}^{u_{1}}\dd u_{2}\int_{0}^{\infty}
{\dd x \over
\braces{\bracks{\pars{a - 1}u_{1} + \pars{b - a}u_{2} + 1}x + 1}^{\, 3}}
\\[5mm] & =
\int_{0}^{1}\dd u_{1}\int_{0}^{u_{1}}\dd u_{2}\,
{1 \over \pars{a - 1}u_{1} + \pars{b - a}u_{2} + 1}
\\[5mm] & =
\left.{1 \over b - a}\int_{0}^{1}\dd u_{1}\,
\ln\pars{\bracks{a - 1}u_{1} + \bracks{b - a}u_{2} + 1}
\,\right\vert_{\ u_{2}\ =\ 0}^{\ u_{2} =\ u_{1}}
\\[5mm] & =
{1 \over b - a}\int_{0}^{1}\dd u_{1}\braces{\vphantom{\Large A}%
\ln\pars{\bracks{b - 1}u_{1} + 1} -
\ln\pars{\bracks{a - 1}u_{1} + 1}}
\\[5mm] & =
{1 \over b - a}\braces{%
\bracks{-1 + {b\ln\pars{b} \over b - 1}} -
\bracks{-1 + {a\ln\pars{a} \over a - 1}}}
\\[5mm] & =
\bbx{{\pars{b - 1}a\ln\pars{a} - \pars{a - 1}b\ln\pars{b} \over
\pars{a - 1}\pars{a - b}\pars{b - 1}}}
\end{align}
A: Repeating the steps you probably did
$$\frac{1}{(x+1)(ax+1)(bx+1)}=\frac{a^2}{(a-1) (a-b) (a x+1)}+\frac{b^2}{(b-1) (b-a) (b x+1)}+\frac{1}{(a-1) (b-1) (x+1)}$$
Now, assuming $a>0$, $b>0$ and $a\neq b$, consider
$$I(p)=\int_0^p\frac{dx}{(x+1)(ax+1)(bx+1)} $$
$$I(p)=\frac{(a-b) \log (p+1)+a (b-1) \log (a p+1)-(a-1) b \log (b p+1)}{(a-1) (b-1)
   (a-b)}$$
Now, let $p=\frac 1 t$ and use Taylor series to get
$$I\left(\frac 1t\right)=\frac{a (b-1) \log (a)-b(a-1) \log (b)}{(a-1) (b-1) (a-b)}-\frac{t^2}{2 a   b}+O\left(t^3\right)$$ and, back to $p$
$$I(p)=\frac{a (b-1) \log (a)-b(a-1) \log (b)}{(a-1) (b-1) (a-b)}-\frac{1}{2 a   bp^2}+O\left(\frac{1}{p^3}\right)$$
A: The estimation of the $\lim_{\infty}$ part is more subtle that you seem to have thought. 
Remove the useless absolute values, take the divergent part wrt $x$ (ie some $\ln{x}$) apart from the rest and enjoy seeing its coefficient vanish. The rest will be a linear combination of logarithms of convergent functions of $x$ ($1+1/x$, $b+1/x$, and so on). 
A: As pointed out by other user, you must take cancellations into consideration when evaluating the limit $x\to\infty$. Hints for such cancellation can be glimpsed from the fact that (1) $\log(px+q) \sim \log x$ as $x\to\infty$ for $p > 0$ and that (2) the sum of `coefficients' is zero: $(a-1)b + (a-ab) + (b-a) = 0$.
So here is an actual computation: Using the fact that
$$(a-1)b \log x + (a-ab) \log x + (b-a) \log x = 0, $$
we find that
\begin{align*}
&(a-1)b \log(bx+1) + (a-ab) \log(ax+1) + (b-a) \log (x+1) \\
&=(a-1)b [\log(bx+1) - \log x] + (a-ab) [\log(ax+1) - \log x] + (b-a) [\log (x+1) - \log x] \\
&= (a-1)b \log\left(b+\frac{1}{x}\right) + (a-ab) \log\left(a+\frac{1}{x}\right) + (b-a) \log \left(1+\frac{1}{x}\right) \\
&\xrightarrow[x\to\infty]{} (a-1)b \log b + (a-ab) \log a + \underbrace{(b-a) \log 1}_{=0}
\end{align*}
