Answer
$\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\frac{1+4\cot x}{4-\cot x}dx}=\ln \left( \frac{4\sqrt{2}}{3} \right)
$
Work Step by Step
$I=\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\frac{1+4\cot x}{4-\cot x}dx}\\
I=\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\frac{17}{4-\cot x}-4dx}\\
I=\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\frac{17}{4-\cot x}dx}-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{4dx}\\
I=\displaystyle\color{Plum}{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\frac{17}{4-\cot x}dx}}\color{LimeGreen}{-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{4dx}}\\$
$\displaystyle\color{LimeGreen}{-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{4dx}\\
-\left[ 4x \right] _{\frac{\pi}{4}}^{\frac{\pi}{2}}}$
$\displaystyle\color{Plum}{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\frac{17}{4-\cot x}dx}}
$
$\displaystyle{\left[\begin{matrix}
b=\cot x& dx=-\frac{db}{\cos\text{ec}^2x}& 1+b^2\\
\end{matrix}=\cot ^2x+1\right]}$
$\displaystyle\int_1^0{\frac{17}{4-b}\times -\frac{db}{b^2+1}}\\
\displaystyle-17\int_1^0{\frac{1}{\left( 4-b \right) \left( b^2+1 \right)}db}\\
\displaystyle-17\int_1^0{\frac{1}{-\left( b-4 \right) \left( b^2+1 \right)}db}\\
\displaystyle17\int_1^0{\frac{1}{\left( b-4 \right) \left( b^2+1 \right)}db}$
$\displaystyle\frac{1}{\left( b-4 \right) \left( b^2+1 \right)}\equiv \frac{A}{b-4}+\frac{Bb+C}{b^2+1}\\
1=A\left( b^2+1 \right) +\left( b-4 \right) \left( Bb+C \right) \\
1=Ab^2+A+Bb^2+Cb-4Bb-4C\\
1=\left( A+B \right) b^2+\left( C-4B \right) b+\left( A-4C \right)$
$b=0\ \ \therefore \ \ 1=A-4C\ \cdots\ \left( 1 \right) \\
\begin{eqnarray}
b=1\ \ \therefore \ \ \quad0&=&A+B+C-4B\\
3B&=&A+C\ \cdots\ \left( 2 \right)
\end{eqnarray}$
$\begin{eqnarray}
b=-1\ \therefore \ 0&=&A+B-C+4B\\
-5B&=&A-C\ \cdots\ \left( 3 \right)
\end{eqnarray}$
$\left( 2 \right) +\left( 3 \right) \\
\begin{eqnarray}
-2B&=&2A\\
-B&=&A\ ...\ \left( 4 \right) \end{eqnarray}$
$\left( 4 \right)$ in $\left( 3 \right)\\
\begin{eqnarray}
-5B&=&-B-C\\
4B&=&C
\end{eqnarray}$
$\left( 4 \right) ,\left( 5 \right)$ in $\left( 1 \right) \\
\begin{eqnarray}
1&=&A-4C\\
1&=&-B-16B\\
B&=&-1/17\\
\end{eqnarray}\\
A=1/17\\
C=-4/17\\
$
$\displaystyle17\int_1^0{\frac{1}{17\left( b-4 \right)}+\frac{-\frac{b}{17}-\frac{4}{17}}{b^2+1}db}\\
\displaystyle\int_1^0{\frac{1}{\left( b-4 \right)}-\frac{b-4}{b^2+1}db}\\
\displaystyle\int_1^0{\frac{1}{b-4}-\frac{b}{b^2+1}+\frac{4}{b^2+1}db}\\
\displaystyle\color{plum}{\int_1^0{\frac{1}{b-4}}db-\int_1^0{\frac{b}{b^2+1}db}+4\int_1^0{\frac{1}{b^2+1}db}}\\
\displaystyle\color{orange}{\int_1^0{\frac{1}{b-4}}db}\color{tan}{-\int_1^0{\frac{b}{b^2+1}db}}\color{blue}{+4\int_1^0{\frac{1}{b^2+1}db}}$
$\displaystyle\color{orange}{\int_1^0{\frac{1}{b-4}}\\
\left[ \ln \left| b-4 \right| \right] _{1}^{0}}$
$\displaystyle\color{tan}{-\int_1^0{\frac{b}{b^2+1}db}}$
$\displaystyle{\left[\begin{matrix}
z=b^2+1& db=\frac{dz}{2b}\\
\end{matrix}\right]}
$
$
\displaystyle-\int_2^1{\frac{b}{z}\frac{dz}{2b}}\\
\displaystyle-\frac{1}{2}\int_2^1{\frac{1}{z}dz}\\
\displaystyle\color{tan}{-\frac{1}{2}\left[ \ln z \right] _{2}^{1}}$
$\displaystyle\color{blue}{4\int_1^0{\frac{1}{b^2+1}db}}$
$\displaystyle{\left[\begin{matrix}
b=\tan \theta& b^2=\tan ^2\theta\\
\frac{db}{d\theta}=\sec ^2\theta& db=\sec ^2\theta d\theta\\
\end{matrix}\right]}$
$\displaystyle4\int_{\frac{\pi}{4}}^0{\frac{1}{\sec ^2\theta}\sec ^2\theta d\theta}\\
\displaystyle4\int_{\frac{\pi}{4}}^0{1\ d\theta}\\
\displaystyle\color{blue}{{4\left[ \theta \right] _{\frac{\pi}{4}}^{0}}}$
$I=\displaystyle\color{orange}{\left[ \ln \left| b-4 \right| \right] _{1}^{0}}\color{tan}{-\frac{1}{2}\left[ \ln z \right] _{2}^{1}}\color{blue}{+4\left[ \theta \right] _{\frac{\pi}{4}}^{0}}\color{limegreen}{-\left[ 4x \right] _{\frac{\pi}{4}}^{\frac{\pi}{2}}}\\
I=\displaystyle\ln \left( \frac{4\sqrt{2}}{3} \right)
$
