I assume you mean Integration by Parts.
The general formula is:
#color(green)(int udv = uv - intvdu)#
Let's suppose you were integrating #arctanx#.
#int arctanxdx = int 1*arctanxdx#
You pick a #u# and a #dv#. It's a good idea with this integration technique to pick a #u# that you can easily differentiate, and a #dv# that you can easily antidifferentiate.
#1# is trivial to antidifferentiate, and at this point you would have learned the derivative of #arctanx#, which is #1/(1+x^2)#.
(If you had picked the other way around, it would have been counterproductive---why should you know the integral of something that you are currently integrating, right?)
Let:
#u = arctanx#
#dv = 1dx#
#v = x#
#du = 1/(1+x^2)dx#
Then just plug it in and solve.
#=> xarctanx - intx/(1+x^2)dx#
Notice how #d/(dx)[1+x^2] = d/(dx)[x^2] = 2x#. You can get #2x# into the integral and then undo it to create a substitution.
Let #u = 1+x^2#, and #du = 2xdx#:
#= xarctanx - 1/2 int (2x)/(1+x^2)dx#
#= xarctanx - 1/2 int 1/udu#
#= xarctanx - 1/2 ln|u|#
#= color(blue)(xarctanx - 1/2 ln(1+x^2) + C)#
