$\int{ \frac{ 4x }{ {\left( x-4 \right)}^{2} } } \mathrm{d} x$
Use the substitution $t=x-4$ to transform the integral$\int{ \frac{ 4t+16 }{ {t}^{2} } } \mathrm{d} t$
Separate the fraction into $2$ fractions$\int{ \frac{ 4t }{ {t}^{2} }+\frac{ 16 }{ {t}^{2} } } \mathrm{d} t$
Cancel out the common factor $t$$\int{ \frac{ 4 }{ t }+\frac{ 16 }{ {t}^{2} } } \mathrm{d} t$
Use the property of integral $\int{ f\left( x \right)\pmg\left( x \right) } \mathrm{d} x=\int{ f\left( x \right) } \mathrm{d} x\pm\int{ g\left( x \right) } \mathrm{d} x$$\int{ \frac{ 4 }{ t } } \mathrm{d} t+\int{ \frac{ 16 }{ {t}^{2} } } \mathrm{d} t$
Use $\int{ \frac{ a }{ x } } \mathrm{d} x=a \times \ln\left({|x|}\right)$ to evaluate the integral$4\ln\left({|t|}\right)+\int{ \frac{ 16 }{ {t}^{2} } } \mathrm{d} t$
Evaluate the indefinite integral$4\ln\left({|t|}\right)-\frac{ 16 }{ t }$
Substitute back $t=x-4$$4\ln\left({|x-4|}\right)-\frac{ 16 }{ x-4 }$
Add the constant of integration $C \in ℝ$$\begin{array} { l }4\ln\left({|x-4|}\right)-\frac{ 16 }{ x-4 }+C,& C \in ℝ\end{array}$