1. ## Indefinite 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

2. Originally Posted by john-1
Can someone show me the steps to finding the integral of (x+1)/(x^2+2x) [ln(x^2+2x)] dx
Do you mean 'show that' $\displaystyle \int \frac{x+1}{x^2+2x}~dx = \ln(x^2+2x)$ or 'find' $\displaystyle \int \frac{x+1}{x^2+2x} \ln(x^2+2x)~dx$

??

3. Originally Posted by pickslides
Do you mean 'show that' $\displaystyle \int \frac{x+1}{x^2+2x}~dx = \ln(x^2+2x)$ or 'find' $\displaystyle \int \frac{x+1}{x^2+2x} \ln(x^2+2x)~dx$

??

4. The second one, the first on is not so true...

in second integral use integration by parts...

5. could you show me the steps please? I only have the answer with me in my text but no steps

6. 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 $\displaystyle \displaystyle \int\frac{(x+1)\ln(x^2+2x)}{x^2+2x}\;{dx} = \frac{1}{4}\ln^2(x^2+2x)+k$.

7. TheCoffeeMachine, in your second step, why did you put (x^2 + 2x) / (x+1)? Where did that come from? mostly confused about 2nd step

8. 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!