Can someone show me the steps to finding the integral of (x+1)/(x^2+2x) [ln(x^2+2x)] dx

I'm not sure where to begin because I don't have any integration rules with the lnx

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- Aug 4th 2010, 08:07 PMjohn-1Indefinite Integrals
Can someone show me the steps to finding the integral of (x+1)/(x^2+2x) [ln(x^2+2x)] dx

I'm not sure where to begin because I don't have any integration rules with the lnx - Aug 4th 2010, 08:12 PMpickslides
- Aug 4th 2010, 08:14 PMjohn-1
- Aug 4th 2010, 08:17 PMAlso sprach Zarathustra
The second one, the first on is not so true...

in second integral use integration by parts... - Aug 4th 2010, 08:18 PMjohn-1
could you show me the steps please? I only have the answer with me in my text but no steps

- Aug 4th 2010, 08:54 PMTheCoffeeMachine
Using the substitution $\displaystyle u = \ln(x^2+2x)$, we have:

$\displaystyle \displaystyle \int\frac{(x+1)\ln(x^2+2x)}{x^2+2x}\;{dx} = \frac{1}{2}\int\left(\frac{x+1}{x^2+2x}\right)\lef t(\frac{x^2+2x}{x+1}\right)\cdot u \;{du} = \frac{1}{2}\int{u}\;{du} = \frac{1}{4}u^2+k.$

Therefore $\displaystyle \displaystyle \int\frac{(x+1)\ln(x^2+2x)}{x^2+2x}\;{dx} = \frac{1}{4}\ln^2(x^2+2x)+k$. - Aug 5th 2010, 11:31 AMjohn-1
TheCoffeeMachine, in your second step, why did you put (x^2 + 2x) / (x+1)? Where did that come from? mostly confused about 2nd step

- Aug 5th 2010, 11:58 AMAllanCuz
If $\displaystyle u = ln(x^2 + 2x) $ then $\displaystyle \frac{du}{dx} = \frac{2x +2}{x^2+2x} \to dx = (\frac{1}{2}) \frac{x^2 + 2x}{x+1}du$

He then subbed this into the integral for dx!