indefinite integral of rational function I'm asked the question, evaluate:
$$
\int(mx+a)^{-1}dx
$$
I know this has to do with substitution: $\int f(g(x))g'(x)dx=\int f(u)du$. But I cant see how to apply this. This is what I do:
$$
Let \ \ u=mx+a \ \ then \ \ du=mdx
$$
Embarrasing, but, thats about where I stop understanding...

$$\int\left(mx+a\right)^{-1}\space\text{d}x=\int\frac{1}{mx+a}\space\text{d}x=$$

Substitute $u=a+mx$ and $\text{d}u=m\space\text{d}x$:

$$\frac{1}{m}\int\frac{1}{u}\space\text{d}u=\frac{\ln\left|u\right|}{m}+\text{C}=\frac{\ln\left|a+mx\right|}{m}+\text{C}$$
