# Simple Integration

1. Dec 22, 2013

Could anyone please tell me why is the integration of âˆ«(x+2)/(x-1)dx â‰  (x-1)+3ln(x-1).

I got it using substitution of u=x+1.

2. Dec 22, 2013

### vanhees71

I'd do the integral as follows
$$\int \mathrm{d} x \frac{x+2}{x-1}=\int \mathrm{d} x \frac{x-1+3}{x-1} = \int \mathrm{d} x \left (1+\frac{3}{x-1} \right )=x+3 \ln(|x-1|)+C.$$
So up to your missing modulus under the log (which only means that your result is valis for $x>1$ only), you got the correct solution. Why do you think it's wrong?

3. Dec 22, 2013