# How do you find the indefinite integral of int 1/(xlnx^3)?

Dec 12, 2016

$\int \frac{\mathrm{dx}}{x \ln {x}^{3}} = \frac{1}{3} \ln \left(\ln x\right) + C$

#### Explanation:

We begin by exploiting the properties of logarithms and write:

$\int \frac{\mathrm{dx}}{x \ln {x}^{3}} = \int \frac{\mathrm{dx}}{3 x \ln x}$

Now we note that $d \left(\ln x\right) = \frac{\mathrm{dx}}{x}$, so that:

$\int \frac{\mathrm{dx}}{3 x \ln x} = \frac{1}{3} \int \frac{d \left(\ln x\right)}{\ln} x = \frac{1}{3} \ln \left(\ln x\right) + C$

Dec 12, 2016

Substitute $x = {e}^{u}$ to get $\frac{1}{3} \ln \left(\ln\right) x + C$

#### Explanation:

$I = \int \frac{1}{x \ln {x}^{3}} \mathrm{dx} = \frac{1}{3} \int \frac{1}{x \ln x} \mathrm{dx}$ by fundamental law of logarithms.
Now substitute $x = {e}^{u}$, $\ln x = u$ and $\mathrm{dx} = {e}^{u} \mathrm{du}$
to get $I = \frac{1}{3} \int \frac{1}{{e}^{u} u} {e}^{u} \mathrm{du} = \frac{1}{3} \int \frac{1}{u} \mathrm{du} = \frac{1}{3} \ln x + C$.
(Or just substitute $x = {e}^{u}$ at the start.)

Dec 12, 2016

You may have also meant to type $\int \frac{1}{x {\left(\ln x\right)}^{3}} \mathrm{dx}$. If this is the case, look here; if not, you can still learn something from this!

If we do indeed have $\int \frac{1}{x {\left(\ln x\right)}^{3}} \mathrm{dx}$, let $u = \ln x$. This implies that $\mathrm{du} = \frac{1}{x} \mathrm{dx}$.

Then we have the equivalent integrals $\int \frac{1}{\ln x} ^ 3 \left(\frac{1}{x} \mathrm{dx}\right) = \int \frac{1}{u} ^ 3 \mathrm{du} = \int {u}^{-} 3 \mathrm{du}$.

Now we can use $\int {u}^{n} \mathrm{du} = {u}^{n + 1} / \left(n + 1\right) + C$, where $n \ne - 1$, which is not an issue here.

The integral then becomes ${\left(\ln x\right)}^{-} \frac{2}{- 2} + C$, or $\frac{- 1}{2 {\left(\ln x\right)}^{2}} + C$.