# What is the antiderivative of 1/[x(ln(x^3))]?

Jan 20, 2016

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

Note that $\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$ and,

more importantly,

$\frac{d}{\mathrm{dx}} \left(\ln \left({x}^{3}\right)\right) = \frac{3}{x}$.
(You can get this by the chain rule, or more simply, by noting that $\ln \left({x}^{3}\right) = 3 \ln x$.)

This integral can be evaluated by substitution:

Let $u = \ln \left({x}^{3}\right)$, then $\mathrm{du} = \frac{3}{x} \mathrm{dx}$, so $\frac{1}{x} \mathrm{dx} = \frac{1}{3} \mathrm{du}$

Upon substitution, the integral becomes:

$\int \frac{1}{3} {u}^{-} 1 \mathrm{du} = \frac{1}{3} \ln \left\mid u \right\mid + C$.

Therefore:

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

Another way to proceed:

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

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

Let $u = \ln x$ and proceed to get:

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

It looks different, but what is the difference?

$\frac{1}{3} \ln \left\mid \ln \right\mid \left({x}^{3}\right) = \frac{1}{3} \ln \left\mid 3 \ln x \right\mid$

$= \frac{1}{3} \left(\ln 3 + \ln \left\mid \ln \right\mid x\right) + C$

$= \frac{1}{3} \ln 3 + \frac{1}{3} \ln \left\mid \ln \right\mid x + C$

So the difference between the expressions is a constant.
The two general answers simply have different $C$'s for particular answers.