Integrating $\int \frac{1}{x(\ln x)^2}\,dx$

Question

I would like to know how to integrate $$\int \frac{1}{x(\ln x)^2}\,dx$$

Specifically, I would like to know how to do this by substituting for $(\ln x)^2$. Thank you!

Answer 1

Let $u=\ln x$. Then $du=\dfrac1x\, dx$. Substituting, we get $$\int\frac{1}{x(\ln x)^2}\,dx=\int\frac{1}{u^2}\,du.$$

Can you take it from here?

Answer 2

Substituting for $(\ln x)^2$ is certainly not where I would start, but okay...

Let $u = (\ln x)^2$ so that \begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x} &= 2 (\ln x) \cdot \frac{1}{x} \\ &= \frac{2 \sqrt{u}}{x} \\ \frac{1}{2\sqrt{u}} \frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{1}{x} \text{.} \end{align*} Then \begin{align*} \int \frac{1}{x (\ln x)^2} \,\mathrm{d}x &= \int \frac{1}{u} \cdot \frac{1}{2\sqrt{u}} \,\mathrm{d}u \\ &= \frac{1}{2} \int u^{-3/2} \,\mathrm{d}u \\ &= \frac{1}{2} \left( \frac{u^{-1/2}}{-1/2} + C \right) \\ &= -u^{-1/2} + C \\ &= -\left( (\ln x)^2 \right)^{-1/2} + C \\ &= -(\ln x)^{-1} + C \end{align*}