Question

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` ?

Answer 1

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`