Indefinite Integrals

Sep 2009
78
0
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
Sep 2008
5,237
1,625
Melbourne
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' \(\displaystyle \int \frac{x+1}{x^2+2x} \ln(x^2+2x)~dx \)

??
 
Sep 2009
78
0
Do you mean 'show that' \(\displaystyle \int \frac{x+1}{x^2+2x}~dx = \ln(x^2+2x) \) or 'find' \(\displaystyle \int \frac{x+1}{x^2+2x} \ln(x^2+2x)~dx \)

??
the second one please
 
Dec 2009
1,506
434
Russia
The second one, the first on is not so true...

in second integral use integration by parts...
 
Sep 2009
78
0
could you show me the steps please? I only have the answer with me in my text but no steps
 
Mar 2010
715
381
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 \(\displaystyle \displaystyle \int\frac{(x+1)\ln(x^2+2x)}{x^2+2x}\;{dx} = \frac{1}{4}\ln^2(x^2+2x)+k\).
 
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Sep 2009
78
0
TheCoffeeMachine, in your second step, why did you put (x^2 + 2x) / (x+1)? Where did that come from? mostly confused about 2nd step
 
Apr 2010
384
153
Canada
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!