Logarithmic integration of two related terms How do I prove that
$$\int_{0}^{\infty} du \left(\frac{u^{2}}{(u+a)^{3}} - \frac{u^{2}}{(u+b)^{3}}\right) = \ln \left(\frac{b}{a}\right)?$$
 A: You can use the substitution $t = u + c$
$$ \int_{0}^{+\infty}\frac{u^2}{(u+c)^3}\,du = \int_{c}^{+\infty}\frac{t^2-2tc+c^2}{t^3}\,dt = \int_{c}^{+\infty}\frac{1}{t}\,dt - \int_{c}^{+\infty}\frac{2c}{t^2}\,dt + \int_{c}^{+\infty}\frac{c^2}{t^3}\,dt = \int_{c}^{+\infty}\frac{1}{t}\,dt + 2 - \frac{1}{2}$$
and plug in $a$ and $b$ to obtain
$$\int_{0}^{+\infty} du \left(\frac{u^{2}}{(u+a)^{3}} - \frac{u^{2}}{(u+b)^{3}}\right) = \int_{a}^{+\infty}\frac{1}{t}\,dt-\int_{b}^{+\infty}\frac{1}{t}\,dt = \ln \left(\frac{b}{a}\right)$$
A: By integration by parts it boils down to
$$ \int_{0}^{+\infty}\left(\frac{1}{u+a}-\frac{1}{u+b}\right)\,du = \log(b)-\log(a).$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{0}^{\infty}\dd u \bracks{{u^{2} \over \pars{u + a}^{3}} -
{u^{2} \over \pars{u + b}^{3}}}
\\[5mm] = &\ \
\int_{0}^{\infty}
\bracks{{u^{2} \over \pars{u + a}^{3}} - {1 \over u + a}}\dd u -
\int_{0}^{\infty}
\bracks{{u^{2} \over \pars{u + b}^{3}} - {1 \over u + b}}\dd u
+
\int_{0}^{\infty}\pars{{1 \over u + a} - {1 \over u + b}}\dd u =
\\[5mm] = &\ \
\braces{\int_{0}^{\infty}
\bracks{{u^{2} \over \pars{u + 1}^{3}} - {1 \over u + 1}}\dd u -
\int_{0}^{\infty}
\bracks{{u^{2} \over \pars{u + 1}^{3}} - {1 \over u + 1}}\dd u}
+
\left.\ln\pars{u + a \over u + b}\right\vert_{\ u\ =\ 0}^{\ u\ \to\ \infty}
\\[5mm] = &\
\bbx{\ds{\ln\pars{b \over a}}}
\end{align}
