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

Compute

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

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

• Where would you get log 0 ? – Shailesh Jun 6 '18 at 1:19
• I was evaluating the limits using u instead of ax+b. But I guess I had to change back to ax+b as caverac showed me. What do you think? – Felipe Camargo Jun 6 '18 at 1:34
• If you make a substitution, then you need to change the limits too. If $x \in [0,1]$, then $u \in [b, a+b]$. – Théophile Jun 6 '18 at 1:38
• If a change back before, as caverac did in his solution, is wrong? – Felipe Camargo Jun 6 '18 at 1:55

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

• That's perfect, Phil! Thanks – Felipe Camargo Jun 6 '18 at 2:38

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

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

• Thanks caverac. It's not wrong to change back the u part before apply the definite integral part, since I made the integral with u? – Felipe Camargo Jun 6 '18 at 1:29