Integration by Parts the Adult Way

A lot of people do integration by parts by defining the variables $\displaystyle u=...$ and $\displaystyle v'=...$ and flip them around.

There is a more adult way of doing this, which looks nicer and is a lot faster.

Say we have the integral, $\displaystyle \int xe^x dx$

The idea is to turn one of the factors into a derivative. For example, we know that $\displaystyle (e^x)' = e^x$.

Thus, we can think of the integral as,

$\displaystyle \int x \left( e^x \right) ' dx$

The next step is to take the function __inside__ the differenciation operator and multiply it with the function unaffected with the differenciation and multiply them together. That is the $\displaystyle uv$ part that you get.

Thus, we get $\displaystyle xe^x - \int ... $

The next step is to take the derivative of the function which was unaffected by differenciation and multiply it by the function inside the differenciation sign. This is our $\displaystyle u'v$ part.

In this case we get, $\displaystyle xe^x - \int (x) ' e^x dx = xe^x - e^x + C$.

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Here is another example,

$\displaystyle \int \ln x dx = \int \ln x \left( x \right) ' dx = x\ln x - \int 1 dx = x\ln x - x + C$.

Look how fast that is.

Here is another example,

$\displaystyle \int x^2 \sin 2x dx = \int x^2 \left( -\frac{1}{2} \cos 2x \right) ' dx = -\frac{1}{2}x^2 \cos 2x + \int x\cos 2x dx $

$\displaystyle =-\frac{1}{2}x^2\cos 2x + \int x \left( \frac{1}{2} \sin 2x \right)' dx = - \frac{1}{2}x^2\cos 2x + \frac{1}{2}x\sin 2x - \int \frac{1}{2}\sin 2x dx = $$\displaystyle -\frac{1}{2}x^2 \cos 2x + \frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x + C$

My point is that it is a lot easier to keep track of everything doing integration this way. Because you do not need to go out of your way to write $\displaystyle u$ and $\displaystyle v'$.