Evaluate the integral: $\int_{0}^{1} \frac{1}{ax+b} dx$

Question

Compute

$\int_{0}^{1} \frac{1}{ax+b} dx$

I tried with substituion method but got stuck in log(0). Can someone help me?

Answer 1

I like the simplicity of $\int \frac{{\rm d}x}{a x + b} = \frac{1}{a}\int\frac{{\rm d}x}{x+\frac{b}{a}} = \frac{1}{a}\ln ( x + \frac{b}{a})$

Answer 2

Call $u = a x + b$, then ${\rm d}u = a{\rm d}x$ and

$$ \int \frac{{\rm d}x}{a x + b} = \frac{1}{a}\int\frac{{\rm d}u}{u} = \frac{1}{a}\ln u = \frac{1}{a}\ln(a x + b) $$

So that

$$ \int_0^1 \frac{{\rm d}x}{a x + b} = \frac{1}{a}[\ln (a + b) - \ln b] = \frac{1}{a}\ln\left(\frac{a}{b} + 1\right) $$