# Question about this integral

1. Mar 15, 2007

### alba_ei

1. The problem statement, all variables and given/known data

$$\int \frac{ae^\theta+b}{ae^\theta-b} \, d\theta$$

3. The attempt at a solution

i took $$u = ae^\theta-b$$ so $$e^\theta = \frac{u + b}{a}$$ then i substituded back into the integral and iget this

$$\int \frac{u + b + b}{u} \, du$$

$$\int du +\int \frac{2b}{u} \, du$$

$$= u \du + 2b \ln u +C$$

$$= u + 2b \ln u +C$$

$$= ae^\theta-b + 2b\ln (ae^\theta-b)$$

but the answer of the book is
$$\int \frac{ae^\theta+b}{ae^\theta-b} \, d\theta = 2\ln (ae^\theta-b) - \theta + C$$
what did i do wrong?

Last edited: Mar 15, 2007
2. Mar 15, 2007

### Galileo

You didn't subsitute properly. You have to change dtheta too.

3. Mar 15, 2007

### ziad1985

write d0 as what it should equal to du
for example if u=x^2
du=2*xdx

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