Question

What is integration by parts and why does it work?

Answer 1

Integration by parts is a method of calculating integrals (both definite and indefinite).

Explanation:

Integration by parts is based on a formula:

`intf'(x)g(x)dx=f(x)*g(x)-intf(x)g'(x)dx`.

It is a method to change one integral to another which may be easier to calculate.

This method is based on the formula for the derivative of ptroduct of 2 functions:

`[f(x)*g(x)]'=f'(x)*g(x)+f(x)*g'(x)`

`f'(x)*g(x)=[f(x)*g(x)]'-f(x)*g'(x)`

Now if we integrate both sides of the equation we get:

`intf'(x)*g(x)dx=f(x)*g(x)-intf(x)*g'(x)dx`

As an example I will calculate `intlnxdx`

`intlnxdx=int(x')*lnxdx=|(u=x,v=lnx),(u'=1,v'=1/x)|`

`=xlnx-intx*(lnx)'dx=xlnx-intx*1/xdx=xlnx-int1dx=xlnx-x+C`

In the example we changed an integral of `f(x)=lnx` to an easier integral of `g(x)=1`

Answer 2

It works because of the product rule for derivatives and because integration is the inverse of differentiation.

Explanation:

The product rule tells us that

`d/dx(uv) = [(du)/dx]v+u[(dv)/dx]`.

Using differentials, this says that:

`d(uv) = [du]v+u[dv]`,

more conveniently written

`d(uv) = vdu+udv`.

Now, Integrate both sides:

`int d(uv) = int(vdu+udv)`
.
So,

`int d(uv) =int vdu+ int udv`.

By the inverse relationship of differentiation and integration, we can simplify the left side to get:

`uv =int vdu+ int udv`.

`int vdu+ int udv = uv`.

And finally, subtracting `int vdu` from both sideswe get

`int udv = uv - intvdu`.

Alternative Notations

Rather than using the dufferential notation shown above, others prefer to use:

`int f(x)g'(x)dx = f(x)g(x)-intg(x)f'(x)dx`.

Still others write

`int f(x)g(x)dx = f(x)intg(x)dx - int [d/dx(f(x))intg(x)dx]`

(Or something like that. I've seen the last one rarely.)