# Thread: Integration by parts confusion

1. ## Integration by parts confusion

Hello,

I have a pretty simple integration by parts question I'm supposed to solve, but for some reason I can't seem to get the right answer...Can someone point out where I've gone wrong?

$\displaystyle \int x\, ln x \, dx$

So, $\displaystyle u = ln x \,$

$\displaystyle dv= x \, dx \,$

$\displaystyle du= \frac {1}{x} \,dx$

$\displaystyle v = \frac {x^2}{2}$

Our tutor instructed us that if there is a natural log in the question, that that should be set as the variable to be deriviated, as we haven't been taught a technique for finding the integral of ln x.

Anyway, I worked the question out as follows...

$\displaystyle = uv - \int v \, du$

$\displaystyle = ln\,x (\frac {x^2}{2}) - \int \frac {x^2}{2} \, \frac {1}{x} \, dx$

$\displaystyle = \frac {x^2 ln \,x}{2} - \frac {1}{2} \int x^2 \, \frac {1}{x} \,dx$

$\displaystyle = \frac {x^2 ln \,x}{2} - \frac {1}{2} \cdot \frac {x^3}{3} \, ln\,x\,+\,C$

$\displaystyle = \frac {x^2 ln \,x}{2} - \frac {x^3 \,ln\,x}{6}\,+\,C$

Apparently, I'm supposed to be getting this though...

$\displaystyle = \frac {x^2 ln \,x}{2} - \frac {x^2}{4} \,+\,C$

Where exactly am I going wrong?

2. ## Re: Integration by parts confusion

$\displaystyle \int x^2\frac{1}{x}dx$ you integrated using a nonexisting rule $\displaystyle \int f\cdot g=\int f \cdot \int g$.
You shoud be doing $\displaystyle \int x^2\frac{1}{x}dx=\int x dx=\frac{x^2}{2}$

3. ## Re: Integration by parts confusion

Originally Posted by astuart
Hello,

I have a pretty simple integration by parts question I'm supposed to solve, but for some reason I can't seem to get the right answer...Can someone point out where I've gone wrong?

$\displaystyle \int x\, ln x \, dx$

So, $\displaystyle u = ln x \,$

$\displaystyle dv= x \, dx \,$

$\displaystyle du= \frac {1}{x} \,dx$

$\displaystyle v = \frac {x^2}{2}$

Our tutor instructed us that if there is a natural log in the question, that that should be set as the variable to be deriviated, as we haven't been taught a technique for finding the integral of ln x.

Anyway, I worked the question out as follows...

$\displaystyle = uv - \int v \, du$

$\displaystyle = ln\,x (\frac {x^2}{2}) - \int \frac {x^2}{2} \, \frac {1}{x} \, dx$

$\displaystyle = \frac {x^2 ln \,x}{2} - \frac {1}{2} \int x^2 \, \frac {1}{x} \,dx$

$\displaystyle = \frac {x^2 ln \,x}{2} - \frac {1}{2} \cdot \frac {x^3}{3} \, ln\,x\,+\,C$
$\displaystyle \int f(x)g(x)dx$ is NOT "$\displaystyle \int f(x)dx\int g(x)dx$"! Instead just note that $\displaystyle \frac{x^2}{2}\frac{1}{x}= \frac{x}{2}$.

$\displaystyle = \frac {x^2 ln \,x}{2} - \frac {x^3 \,ln\,x}{6}\,+\,C$

Apparently, I'm supposed to be getting this though...

$\displaystyle = \frac {x^2 ln \,x}{2} - \frac {x^2}{4} \,+\,C$

Where exactly am I going wrong?

4. ## Re: Integration by parts confusion

Sorry, but I don't quite understand why the integral sign is there then? Are you essentially saying that the integral sign should be ignored, and I should instead multiply $\displaystyle \frac {x^2}{2} \cdot \frac {1}{x}$ resulting in $\displaystyle \frac {x}{2}$??

I'm missing something really simple here, but it just hasn't clicked...

5. ## Re: Integration by parts confusion

Originally Posted by astuart
Sorry, but I don't quite understand why the integral sign is there then? Are you essentially saying that the integral sign should be ignored, and I should instead multiply $\displaystyle \frac {x^2}{2} \cdot \frac {1}{x}$ resulting in $\displaystyle \frac {x}{2}$??

I'm missing something really simple here, but it just hasn't clicked...
Nevermind, I see what you mean now....

$\displaystyle \frac {x^2\,lnx}{2} - \int \frac{x}{2}$ (from multiplying the two terms BEFORE taking the integral)

$\displaystyle = \frac {x^2\,lnx}{2} - \frac {1}{2} \int x$

$\displaystyle = \frac {x^2\,lnx}{2} - \frac {1}{2} \cdot \frac{x^2}{2} \,+\,C$

$\displaystyle = \frac {x^2\,lnx}{2} - \frac {x^2}{4}\,+\,C$

Thank you very much, both of you!!