Integrate algebraic fraction with constant on top?

Question

I understand that if you have $\int \frac{1}{x + 1} dx$ you simply do $\ln(x + 1) + C$. Now I'm slight confused because in my text book, $\int \frac{31}{x - 4} dx$ evaluates to $31\ln(x - 4)$ but $\int\frac{-2}{4x - 1} dx$ becomes $-\frac{1}{2}\ln(4x - 1)$. Why is this?

Answer 1

It is because in $\frac{-2}{4x - 1}$, there is a $4$ in the denominator. We get $$\int \frac{-2}{4x - 1} dx = -2\int \frac{1}{4x - 1} dx = -2\int \frac{1}{4} \frac{1}{u} du = -\frac12 \ln(u) = -\frac12 \ln(4x-1)+C$$

where $u=4x-1$.

Answer 2

When you have $\frac{-2}{4x-1}$ take $y = 4x-1$ so you have $dy=4dx.$ now, your integral is $\frac{-2}{y}\frac{dy}{4}= -\frac{1}{2} \frac{dy}{y} = -\frac{1}{2} \ln(y)$.