How do you use integration by parts to establish the reduction formula #int(ln(x))^n dx = x(ln(x))^n-nint(ln(x))^(n-1)dx# ?

1 Answer
Aug 18, 2014

Remember that Integration by Parts involves the following:

#intudv = uv - intvdu#

To accomplish this, we must choose a term for #u#, and another for #dv#. To figure out which terms would work best, we can use the ILATE method for Integration by Parts:

I nverse
L ogarithm
A lgebraic
T rig
E xponential

This order of priority can help you decide which term should be our #u#, and which should be our #dv#. Whichever term in our equation that's higher on this list should be our #u#. For this particular equation, you can ignore the #n# and treat the #lnx# on its own, making #(lnx)^n# our #u#, and #dx# our #dv#.

Now we must differentiate our #u# and integrate our #dv#, giving us:

#d/dx(lnx)^n = n(lnx)^(n-1)(1/x)#,

and

#int dx = x#

With our original Integration by Parts formula, that gives us:

#int (lnx)^ndx = x(lnx)^n - nint(lnx)^(n-1)(1/x)x#

Since the #x# and the #1/x# cancel out, we end up with the final reduction formula answer:

#int(lnx)^ndx = x(lnx)^n - nint(lnx)^(n-1)dx#