Question

Evaluate (i) \(\int \frac{\mathrm{dx}}{\mathrm{x} \ln \mathrm{x}}\) (ii) \(\int \frac{\mathrm{dx}}{\mathrm{x} \ln \mathrm{x} \ln \ln \mathrm{x}}\)

Short answer

The solution to the integral \(\int \frac{\mathrm{dx}}{\mathrm{x} \ln \mathrm{x} \ln \ln \mathrm{x}}\) is \(- \frac{1}{\ln(\ln x)} + C\), where \(C\) is the constant of integration.

Step-by-step solution

Substitution

Let \(u = \ln x\). Then, \(\frac{du}{dx} = \frac{1}{x}\), which yields \(dx = x\,du\). The integral becomes: $$\int \frac{\mathrm{dx}}{\mathrm{x} \ln \mathrm{x}} = \int \frac{x\,du}{x\,u}.$$

Simplify and Integrate

Now cancelling the common factors of \(x\) in the numerator and denominator, we get: $$\int \frac{x\,du}{x\,u} = \int \frac{1}{u} \mathrm{du}.$$ The integral of \(1/u\) with respect to \(u\) is \(\ln |u|\), therefore the solution to the integral is: \(\ln |u| + C\).

Substitute Back

Finally, we replace \(u\) with \(\ln x\) (initial substitution), so the solution becomes: $$\int \frac{\mathrm{dx}}{\mathrm{x} \ln \mathrm{x}} = \ln |\ln x| + C.$$ (ii) \(\int \frac{\mathrm{dx}}{\mathrm{x} \ln \mathrm{x} \ln \ln \mathrm{x}}\)

Iterative Substitution

In this case, we will perform two substitutions. Let \(v = \ln x\), and then let \(w = \ln v\). Substituting twice, we have the following relationships between the variables: $$x=e^{v}= e^{e^{w}}$$ $$\frac{d v}{dx}=\frac{1}{x} \implies dx = x\, dv = e^{v} dv = e^{e^{w}} dw$$ $$\frac{d w}{dv}=\frac{1}{v} \implies dv = v\, dw = (e^{w}) dw$$ Now the integral becomes: $$\int \frac{\mathrm{dx}}{\mathrm{x} \ln \mathrm{x} \ln \ln \mathrm{x}} = \int \frac{e^{e^{w}} dw}{e^{e^{w}} e^{w} w}.$$

Simplify and Integrate

Now cancelling the common factors of \(e^{e^{w}}\) in the numerator and denominator, we get: $$\int \frac{e^{e^{w}} dw}{e^{e^{w}} e^{w} w} = \int \frac{1}{e^{w} w} \mathrm{dw}.$$ Now, we make the substitution \(u = e^w \implies du = e^w dw\). Then, the integral becomes: $$\int \frac{1}{u w} \mathrm{du} = \int \frac{1}{u^2} \mathrm{du}.$$

Integrate and Substitute Back

The integral of \(1/u^2\) with respect to \(u\) is \(-1/u\), therefore the solution to the integral is: $$- \frac{1}{u} + C = - \frac{1}{e^w} + C = -\frac{1}{\ln v} + C = - \frac{1}{\ln(\ln x)} + C.$$ This yields the final solution: $$\int \frac{\mathrm{dx}}{\mathrm{x} \ln \mathrm{x} \ln \ln \mathrm{x}} = -\frac{1}{\ln(\ln x)} + C.$$