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
Remember that Integration by Parts involves the following:
#intudv = uv - intvdu#
To accomplish this, we must choose a term for
I nverse
L ogarithm
A lgebraic
T rig
E xponential
This order of priority can help you decide which term should be our
Now we must differentiate our
#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
#int(lnx)^ndx = x(lnx)^n - nint(lnx)^(n-1)dx#