Problem source

Find $\displaystyle \ \frac{\d y}{\d x} \ $ if 
$$
y = \frac{ax+b}{cx+d}.
\tag{*}
$$
By using  changes of variable of the form $(*)$, or otherwise,
show that
\[
\int_0^1 \frac{1}{(x+3)^2} \; \ln  \left(\frac{x+1}{x+3}\right)\d x
= {\frac16} \ln3 - {\frac14}\ln 2 - \frac 1{12},
\]
and evaluate the integrals
\[
\int_0^1 \frac{1}{(x+3)^2} \; \ln \left(\frac{x^2+3x+2}{(x+3)^2}\right)\d x
\mbox{    and    }
\int_0^1 \frac{1}{(x+3)^2} \; \ln\left(\frac{x+1}{x+2}\right)\d x
.
\]
[Not on original paper:]
By changing to the variable $y$ defined by
$$  y=\frac{2x-3}{x+1},$$ evaluate the integral 
$$ \int_2^4 \frac{2x-3}{(x+1)^3}\;  \ln\!\left(\frac{2x-3}{x+1}\right)\d x.$$
Evaluate the integral
$$ \int_9^{25} {\big({2z^{-3/2} -5z^{-2}}\big)}\ln{\big(2-5z^{-1/2}\big)}\; \d z.$$

Solution source

\begin{align*}
&& y &= \frac{ax+b}{cx+d} \\
&&&= \frac{\frac{a}{c}(cx+d) - \frac{da}{c} + b}{cx+d} \\
\Rightarrow && y' &= \left (b - \frac{da}{c} \right)(-1)(cx+d)^{-2} \cdot c \\
&&&= (ad-bc)(cx+d)^{-2}
\end{align*}

\begin{align*}
&& y &= \frac{x+1}{x+3} \\
&& \frac{\d y}{\d x} &= \frac{2}{(x+3)^2} \\
\Rightarrow && I &= \int_0^1 \frac{1}{(x+3)^2} \; \ln  \left(\frac{x+1}{x+3}\right)\d x \\
&&&= \int_{y=1/3}^{y=1/2} \frac12 \ln y \, \d y \\
&&&= \frac12 \left [ y \ln y - y \right]_{1/3}^{1/2} \\
&&&= \frac12 \left ( \frac12\ln \frac12 - \frac12 - \frac13 \ln\frac13 + \frac13 \right) \\
&&&= \frac16 \ln 3 -\frac14 \ln 2  -\frac1{12}
\end{align*}

\begin{align*}
&& J &= \int_0^1 \frac1{(x+3)^2} \ln \left ( \frac{x^2+3x+2}{(x+3)^2} \right) \d x \\
&&&= \int_0^1 \frac1{(x+3)^2} \left ( \ln \frac{x+1}{x+3} + \ln \frac{x+2}{x+3} \right) \d x \\
&&&= I + \int_0^1 \frac1{(x+3)^2}  \ln \left ( \frac{x+2}{x+3} \right) \d x \\
&&&= I + \int_{y=2/3}^{y=3/4} \ln y\, \d y \\
&&&= I + \left [  y \ln y- y\right]_{2/3}^{3/4} \\
&&&= I + \left ( \frac34 \ln \frac34 - \frac34 - \frac23 \ln \frac23 + \frac23 \right) \\
&&&= I + \left ( \frac34 \ln 3 - \frac32 \ln 2- \frac1{12} - \frac23 \ln 2 + \frac23 \ln 3\right) \\
&&&= I + \left ( \frac{17}{12} \ln 3 - \frac{13}6 \ln 2- \frac1{12} \right) \\
&&&=  \frac16 \ln 3 -\frac14 \ln 2  -\frac1{12} + \left ( \frac{17}{12} \ln 3 - \frac{13}6 \ln 2- \frac1{12} \right) \\
&&&= \frac{19}{12} \ln 3 -\frac{29}{12}\ln 2 - \frac16
\end{align*}

\begin{align*}
&& K &= \int_0^1 \frac{1}{(x+3)^2} \; \ln\left(\frac{x+1}{x+2}\right)\d x \\
&&&= \int_0^1 \frac{1}{(x+3)^2} \; \left ( \ln\left(\frac{x+1}{x+3}\right) - \ln \left ( \frac{x+3}{x+2} \right) \right)\d x \\
&&&= \frac16 \ln 3 -\frac14 \ln 2  -\frac1{12} - \left ( \frac{17}{12} \ln 3 - \frac{13}6 \ln 2- \frac1{12} \right) \\
&&&= -\frac54 \ln 3 +\frac{23}{12} \ln 2
\end{align*}