# 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 ?

Aug 18, 2014

Remember that Integration by Parts involves the following:

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

To accomplish this, we must choose a term for $u$, and another for $\mathrm{dv}$. 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 $\mathrm{dv}$. 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 $\ln x$ on its own, making ${\left(\ln x\right)}^{n}$ our $u$, and $\mathrm{dx}$ our $\mathrm{dv}$.

Now we must differentiate our $u$ and integrate our $\mathrm{dv}$, giving us:

$\frac{d}{\mathrm{dx}} {\left(\ln x\right)}^{n} = n {\left(\ln x\right)}^{n - 1} \left(\frac{1}{x}\right)$,

and

$\int \mathrm{dx} = x$

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

$\int {\left(\ln x\right)}^{n} \mathrm{dx} = x {\left(\ln x\right)}^{n} - n \int {\left(\ln x\right)}^{n - 1} \left(\frac{1}{x}\right) x$

Since the $x$ and the $\frac{1}{x}$ cancel out, we end up with the final reduction formula answer:

$\int {\left(\ln x\right)}^{n} \mathrm{dx} = x {\left(\ln x\right)}^{n} - n \int {\left(\ln x\right)}^{n - 1} \mathrm{dx}$