# What is int(lnx^9)/x^4 dx?

Nov 8, 2015

It is $9 \left[- \frac{1}{3} {x}^{-} 3 \ln x - \frac{1}{9} {x}^{-} 3 + C\right]$ Which we may prefer to write:
$- \frac{3 \ln x + 1}{x} ^ 3 + C$

#### Explanation:

$\ln {x}^{9} = 9 \ln x$, so

$\int \ln {x}^{9} / {x}^{2} \mathrm{dx} = 9 \int {x}^{- 4} \ln x \mathrm{dx}$ Now use integration by parts.

Let $u = \ln x$ and $\mathrm{dv} = {x}^{-} 4 \mathrm{dx}$,

so $\mathrm{du} = {x}^{- 1} \mathrm{dx}$ and $v = - \frac{1}{3} {x}^{-} 3$

And the integral becomes:

$9 \left[- \frac{1}{3} {x}^{-} 3 \ln x + \frac{1}{3} \int {x}^{-} 3 {x}^{-} 1 \mathrm{dx}\right]$

$= 9 \left[- \frac{1}{3} {x}^{-} 3 \ln x + \frac{1}{3} \int {x}^{-} 4 \mathrm{dx}\right]$

$= 9 \left[- \frac{1}{3} {x}^{-} 3 \ln x - \frac{1}{9} {x}^{-} 3 + C\right]$

Which we may prefer to write:

$= - \frac{3 \ln x + 1}{x} ^ 3 + C$