Integration by parts confusion

**astuart** · October 19th 2012, 04:02 PM

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?

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

So, $u = ln x \,$

$dv= x \, dx \,$

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

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

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...

$= uv - \int v \, du$

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

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

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

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

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

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

Where exactly am I going wrong?

**MathoMan** · October 19th 2012, 04:07 PM

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

**HallsofIvy** · October 19th 2012, 04:09 PM

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?

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

So, $u = ln x \,$

$dv= x \, dx \,$

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

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

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...

$= uv - \int v \, du$

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

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

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

$\int f(x)g(x)dx$ is NOT " $\int f(x)dx\int g(x)dx$ "! Instead just note that $\frac{x^2}{2}\frac{1}{x}= \frac{x}{2}$ .

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

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

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

Where exactly am I going wrong?

**astuart** · October 19th 2012, 04:36 PM

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 $\frac {x^2}{2} \cdot \frac {1}{x}$ resulting in $\frac {x}{2}$ ??

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

**astuart** · October 19th 2012, 04:53 PM

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 $\frac {x^2}{2} \cdot \frac {1}{x}$ resulting in $\frac {x}{2}$ ??

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

Nevermind, I see what you mean now....

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

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

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

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

Thank you very much, both of you!!

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