How to calculate $\int\frac{x}{ax+b}$ dx

Question

How to calculate $\int\frac{x}{ax+b}$ dx ?

I know $\int\frac{1}{ax+b}$ dx but when there is $x$ at the numerator, I don't know where to begin

Answer 1

HINT

We have

$$\frac{x}{ax+b}=\frac1a\frac{ax+b-b}{ax+b}=\frac1a-\frac1a\frac{b}{ax+b}$$

or by parts

$$\int\frac{x}{ax+b}dx=\frac1ax\ln(ax+b)-\frac1a\int\ln(ax+b)dx$$

Answer 2

Rewrite $I=\int\frac x{ax+b} dx$ as $I=\frac 1a\int\frac {ax}{ax+b} dx$

Then $I=\frac 1a\int\frac {ax+b-b}{ax+b} dx=\frac 1a\int \left( \frac {ax+b}{ax+b}- \frac {b}{ax+b}\right) dx =\frac 1a\int \left( 1- \frac {b}{ax+b}\right) dx$

You should be OK from there.