Integral of log(log x) Plus One Over (log x) Squared

This lesson works through a popular Class 12 integration question: integrate $\log(\log x) + \dfrac{1}{(\log x)^2}$. The strategy is to split the integrand into two parts and apply integration by parts to each, then notice that two leftover integrals cancel.

Setting up the split

We want

$I = \int \left( \log(\log x) + \frac{1}{(\log x)^2} \right) dx.$

Split it into two integrals:

$I = \int \log(\log x)\, dx + \int \frac{1}{(\log x)^2}\, dx.$

First part: integrating $\log(\log x)$

Write $\log(\log x) = \log(\log x)\cdot 1$ and use integration by parts with $\log(\log x)$ as the first function and $1$ as the second.

$\int \log(\log x)\, dx = x\log(\log x) - \int \frac{d}{dx}\big(\log(\log x)\big)\cdot x\, dx.$

By the chain rule, $\dfrac{d}{dx}\log(\log x) = \dfrac{1}{\log x}\cdot\dfrac{1}{x}$, so the integrand simplifies:

$\int \frac{1}{\log x}\cdot\frac{1}{x}\cdot x\, dx = \int \frac{1}{\log x}\, dx.$

Therefore

\int \log(\log x)\, dx = x\log(\log x) - \int \frac{1}{\log x}\, dx. \tag{2}

Second part: integrating $\dfrac{1}{\log x}$

Again take $\dfrac{1}{\log x}$ as the first function and $1$ as the second.

$\int \frac{1}{\log x}\, dx = \frac{x}{\log x} - \int \frac{d}{dx}\!\left(\frac{1}{\log x}\right)\cdot x\, dx.$

Since $\dfrac{d}{dx}\left((\log x)^{-1}\right) = -\dfrac{1}{(\log x)^2}\cdot\dfrac{1}{x}$, the leftover term becomes

$-\int x\left(-\frac{1}{(\log x)^2}\cdot\frac{1}{x}\right) dx = +\int \frac{1}{(\log x)^2}\, dx.$

So

\int \frac{1}{\log x}\, dx = \frac{x}{\log x} + \int \frac{1}{(\log x)^2}\, dx. \tag{3}

Combining everything

Substitute $(3)$ into $(2)$, then put that into the original split for $I$:

$I = x\log(\log x) - \left[ \frac{x}{\log x} + \int \frac{1}{(\log x)^2}\, dx \right] + \int \frac{1}{(\log x)^2}\, dx.$

The two copies of $\displaystyle\int \frac{1}{(\log x)^2}\, dx$ have opposite signs and cancel:

$I = x\log(\log x) - \frac{x}{\log x} + c.$

Key takeaways

• Splitting $\int(f+g)\,dx$ into $\int f\,dx + \int g\,dx$ lets you attack each piece separately.
• Integration by parts with $1$ as the second function is the standard trick for integrating a lone awkward function like $\log(\log x)$ or $\tfrac{1}{\log x}$.
• The leftover $\int \tfrac{1}{(\log x)^2}\,dx$ produced by the first part cancels exactly with the second part, giving the clean result $x\log(\log x) - \tfrac{x}{\log x} + c$.