Answer:
[tex]\large \text{$ 5 \ln|4x-15|+C $}[/tex]
Step-by-step explanation:
[tex]\large \displaystyle\begin{aligned}\textsf{let }\:u & =4x-15\\\implies \dfrac{du}{dx} & =4\\\implies dx & =\dfrac{1}{4}\: du\\\end{aligned}[/tex]
[tex]\large\displaystyle\begin{aligned}\int \dfrac{20}{4x-5}\:dx &=\int \dfrac{20}{u}\cdot \dfrac{1}{4}\:du\\\\&=\int \dfrac{20}{4u}\:du\\\\&=\int \dfrac{5}{u}\:du\\\\&=5 \ln |u| +C\\\\&=5 \ln |4x-15| + C\end{aligned}[/tex]
