What is integration by parts and why does it work?

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#

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.)