Evaluate the integral: I = integral dx/(x ln x ln (ln x))
Evaluate the integral: $$ I = \int \frac{dx}{x \ln x \ln (\ln x)}. $$
Explanation of Method
Two nested substitutions strip away the layers of composition: first $u=\ln x$ removes the outer $x$, then $v=\ln u$ removes the remaining logarithm, leaving the elementary integral $\int \frac{dv}{v}$.
- Substitute $u=\ln x$ to convert $\dfrac{dx}{x}$ into $du$.
- Substitute again, $v=\ln u$, to remove the remaining logarithm.
- Integrate using the standard logarithmic antiderivative.
- Substitute back twice to return to $x$.
Step 1 · First substitution
We use the substitution:
$$ u = \ln x \quad \Rightarrow \quad du = \frac{dx}{x}. $$
Rewriting the integral in terms of $u$:
$$ I = \int \frac{du}{u\ln u}. $$
Step 2 · Second substitution
Now, we use another substitution:
$$ v = \ln u \quad \Rightarrow \quad dv = \frac{du}{u}. $$
Thus, the integral simplifies to:
$$ I = \int \frac{dv}{v}. $$
Step 3 · Integrate
The integral of $\dfrac{1}{v}$ is:
$$ I = \ln|v| + C = \ln|\ln u| + C. $$
Step 4 · Substitute back
Substituting back $u = \ln x$:
$$ I = \ln\left|\ln(\ln x)\right| + C. $$
Summary Table
| Substitution | Effect |
|---|---|
| $u = \ln x$ | Converts $dx/x$ into $du$ |
| $v = \ln u$ | Converts $du/u$ into $dv$ |
| Result | $\ln|v| = \ln|\ln(\ln x)|$ |
Additional Notes
- The domain of $I$ requires $x>1$ (so $\ln x>0$ and $\ln(\ln x)$ is defined) and $\ln x \neq 1$ (so $x\neq e$, avoiding division by $\ln(\ln x)=0$).
- This "peeling back the layers" strategy — substituting once for each nested logarithm — generalizes to integrands like $\dfrac{1}{x\ln x\ln(\ln x)\ln(\ln(\ln x))}$, and so on.
- The result can be checked by differentiating $\ln\left|\ln(\ln x)\right|$ with the chain rule (applied twice) and confirming it returns $\dfrac{1}{x\ln x\ln(\ln x)}$.
Result
The solution to the integral is: