1. ## Integration

how do you integrate x-1/x+1 dx from 0 to 1

2. Originally Posted by sheven
how do you integrate x-1/x+1 dx from 0 to 1

Hello,

I assume that you want to calculate $\displaystyle \int^{1}_{0} \frac{x-1}{x+1} dx$ from 0 to 1.

First do a long division: $\displaystyle \frac{x-1}{x+1} =1-\frac{2}{x+1}$.

And now you get: $\displaystyle \int^{1}_{0} \frac{x-1}{x+1} dx=\left[x-2\cdot \ln(x+1) \right]^{1}_{0}$

Put in the borders:

$\displaystyle \left( 1-2\cdot \ln(2) \right)-\left(0-2\cdot \ln(1) \right)=1-2\cdot \ln(2)$

Greetings

EB

3. Instead of doing a long division, remember this trick:

$\displaystyle \frac{{x - 1}}{{x + 1}} = \frac{{x + 1 - 2}}{{x + 1}} = \frac{{x + 1}}{{x + 1}} - \frac{2}{{x + 1}} = 1 - \frac{2}{{x + 1}}$

4. I have one more method
Instead of doing long divison or the trick. You can use substitution,
$\displaystyle u=x+1$, then $\displaystyle u-2=x-1$
Thus, by the substitution rule, (note that $\displaystyle \frac{du}{dx}=1$
$\displaystyle \int\frac{u-2}{u}{du}$
Thus,
$\displaystyle \int u-\frac{2}{u}du$
Thus,
$\displaystyle \frac{1}{2}u^2-2\ln |u| +C$
Substitute back,
$\displaystyle \frac{1}{2}(x+1)^2-2\ln |x+1|+C$.
Q.E.D.