Question

Calculus - need work shown

Answer 1

Easy evaluation!

$5( ln( |4x - 15| ) + c$

Step-by-step explanation:

Feel free to ask if you're left with any doubts.

Answer 2

Answer:

$\large 5 \ln|4x-15|+C$

Step-by-step explanation:

$\large \displaystyle\begin{aligned}\textsf{let }\:u & =4x-15\\\implies \dfrac{du}{dx} & =4\\\implies dx & =\dfrac{1}{4}\: du\\\end{aligned}$

$\large\displaystyle\begin{aligned}\int \dfrac{20}{4x-5}\:dx &=\int \dfrac{20}{u}\cdot \dfrac{1}{4}\:du\\\\&=\int \dfrac{20}{4u}\:du\\\\&=\int \dfrac{5}{u}\:du\\\\&=5 \ln |u| +C\\\\&=5 \ln |4x-15| + C\end{aligned}$