# Indefinite Integrals

#### john-1

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

#### pickslides

MHF Helper
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' $$\displaystyle \int \frac{x+1}{x^2+2x} \ln(x^2+2x)~dx$$

??

#### john-1

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

??

#### Also sprach Zarathustra

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

in second integral use integration by parts...

#### john-1

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

#### TheCoffeeMachine

Using the substitution $$\displaystyle u = \ln(x^2+2x)$$, we have:

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

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

Last edited:
john-1

#### john-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

#### AllanCuz

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!