Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 28

Evaluate $\displaystyle \int^2_1 \frac{(mx)^2}{x^3} dx$ by using Integration by parts.
If we let $u = mx$, then $e^u = x$ and $\displaystyle du = \frac{1}{dx} dx$ we must also make the upper and lower limits in terms of $u$, so...

$
\begin{equation}
\begin{aligned}
\text{so } \int^2_1 \frac{(\ln x)^2}{x^3} dx &= \int^{\ln(2)}_{\ln(1)} \frac{u^2}{(e^u)^3} x du = \int^{\ln 2}_0 \frac{u^2}{(eu)^3} (e^u) du\\
\\
&= \int^{\ln 2}_0 \frac{u^2}{e^{2u}} du
\end{aligned}
\end{equation}
$


To evaluate $\displaystyle \int^{\ln 2}_0 \frac{u^2}{e^{2u}} du$ we must use integration by parts. Then,
if we let $u_1 = u^2$ and $dv_1 = e^{-2u} du$
$du_1 = 2u du$ and $\displaystyle v_1 = \int e^{-2u} du = -\frac{1}{2} e^{-2u}$

So,

$
\begin{equation}
\begin{aligned}
\int^{\ln 2}_0 \frac{u^2}{e^{2u}} du = u_1 v_1 - \int v_1 du_1 &= - \frac{u^2}{2} e^{-2u} - \int -\frac{1}{2} e^{-2u} (2u du)\\
\\
&= -\frac{u^2}{2} e^{-2u} + \int ue^{-2u} du
\end{aligned}
\end{equation}
$

To evaluate $\displaystyle \int ue^{-2u} du$, we must use integration by parts once more, so...
If we let $u_2 = i$ and $dv_2 e^{-2u} du$, then
$du_2 = du$ and $\displaystyle v_2 = \int e^{-2u} du = -\frac{1}{2}e^{-2u}$

So,

$
\begin{equation}
\begin{aligned}
\int ue^{-2u} du = u_2 v_2 - \int v_2 du_2 &= -\frac{u}{2} e^{-2u} - \int \left( -\frac{1}{2}e^{-2u} \right) (du)\\
\\
&= -\frac{u}{2} e^{-2u} + \frac{1}{2} \int e^{-2u} du\\
\\
&= -\frac{u}{2} e^{-2u} + \frac{1}{2} \left( -\frac{1}{2} e^{-2u}\right)\\
\\
&= -\frac{u}{2} e^{-2u} - \frac{1}{4} e^{-2u}
\end{aligned}
\end{equation}
$


Going back to the first equation,

$
\begin{equation}
\begin{aligned}
\int^{\ln 2}_0 \frac{u^2}{e^{2u}} &= -\frac{u^2}{2} e^{-2u} + \left[ -\frac{u}{2} e^{-2u} - \frac{1}{4} e^{-2u} \right]\\
\\
&= -\frac{u^2}{2} e^{-2u} -\frac{u}{2}e^{-2u} - \frac{1}{4} e^{-2u}\\
\\
&= -\frac{e^{-2u}}{2} \left( -u^2 - u - \frac{1}{2} \right)
\end{aligned}
\end{equation}
$


Evaluating from 0 to $\ln 2$

$
\begin{equation}
\begin{aligned}
&= \left[ \frac{e^{-2 \ln 2}}{2} \left( -(\ln2)^2 - \ln 2 - \frac{1}{2} \right) \right] - \left[ \frac{e^{-2(0)}}{2} \left( 0^2 - 0 - \frac{1}{2} \right) \right]\\
\\
&= \frac{3}{16} - \frac{1}{8} \left[ (\ln 2)^2 + \ln 2 \right]
\end{aligned}
\end{equation}
$