Integration

**sheven** · February 22nd 2006, 04:15 PM

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

PLeaSe HeLp!!!!!!!!!!!!!!!

**earboth** · February 22nd 2006, 08:36 PM

Originally Posted by sheven

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

PLeaSe HeLp!!!!!!!!!!!!!!!

Hello,

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

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

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

Put in the borders:

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

Greetings

EB

**TD!** · February 23rd 2006, 10:53 AM

Instead of doing a long division, remember this trick:

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

**ThePerfectHacker** · February 23rd 2006, 12:19 PM

I have one more method

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

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