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#### find_the_fun

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- Feb 1, 2012

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- Feb 1, 2012

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Ok I think I get it. So with integration by parts you choose dv before you calculate v? In other words dv has to be part of the original equation but v does not?We are actually setting:

\(\displaystyle u=\ln(x)\,\therefore\,du=\frac{1}{x}\,dx\)

\(\displaystyle dv=dx\,\therefore\,v=x\)

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Correct...integration by parts states:Ok I think I get it. So with integration by parts you choose dv before you calculate v? In other words dv has to be part of the original equation but v does not?

\(\displaystyle \int u\,dv=uv-\int v\,du\)

So, you want to choose a $u$ and a $dv$, and then from these compute $du$ and $v$.

- Jan 29, 2012

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[tex]\int udv= uv- \int v du[/tex], there is NO "v" in the original integral. You get it by your choice of "dv".

Here, the choice is dv= dx, u= ln(x).

- Mar 1, 2012

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\(\displaystyle \int \ln(x) dx = \int \left(1 \times \ln(x) \right)dx\)

Then you can set \(\displaystyle \frac{dv}{dx} = 1\) and \(\displaystyle u = \ln(x)\)