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Compute

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

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

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    $\begingroup$ Where would you get log 0 ? $\endgroup$ – Shailesh Jun 6 '18 at 1:19
  • $\begingroup$ 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? $\endgroup$ – Felipe Camargo Jun 6 '18 at 1:34
  • $\begingroup$ If you make a substitution, then you need to change the limits too. If $x \in [0,1]$, then $u \in [b, a+b]$. $\endgroup$ – Théophile Jun 6 '18 at 1:38
  • $\begingroup$ If a change back before, as caverac did in his solution, is wrong? $\endgroup$ – Felipe Camargo Jun 6 '18 at 1:55
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I like the simplicity of $\int \frac{{\rm d}x}{a x + b} = \frac{1}{a}\int\frac{{\rm d}x}{x+\frac{b}{a}} = \frac{1}{a}\ln ( x + \frac{b}{a})$

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

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

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  • $\begingroup$ 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? $\endgroup$ – Felipe Camargo Jun 6 '18 at 1:29

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