Problem 19 - Calculus Early Transcendentals

Question

Evaluate the following integrals. Include absolute values only when needed. $$\int_{e^{2}}^{e^{3}} \frac{d x}{x \ln x \ln ^{2}(\ln x)}$$

Step-by-Step Solution

Answer

#Question# Evaluate the integral: $$\int_{e^{2}}^{e^{3}} \frac{1}{x \ln x \ln ^{2}(\ln x)}\,\mathrm{d}x.$$ #Answer# The integral evaluates to: $$\frac{1}{\ln 2} - \frac{1}{\ln 3}.$$

Step 1: Identify a suitable substitution

We are given the integral $$\int_{e^{2}}^{e^{3}} \frac{1}{x \ln x \ln ^{2}(\ln x)}\,\mathrm{d}x.$$ To simplify the expression in the integral, we need to identify a suitable substitution. Since there are nested natural logarithms, let's try a substitution for the innermost logarithm. Let $$u = \ln x.$$ Now, we differentiate both sides with respect to \(x\): $$\frac{d u}{d x} = \frac{1}{x}.$$ So, we can rewrite the integral in terms of \(u\): $$\int_{2}^{3} \frac{1}{u \ln^{2}u}\,\mathrm{d}u.$$

Step 2: Integrate the simplified expression

Now, we want to find the integral $$\int_{2}^{3} \frac{1}{u \ln^{2}u}\,\mathrm{d}u.$$ But we notice that we can perform another substitution to further simplify this integral. Let's try using another substitution for the remaining natural logarithm. Let $$v = \ln u.$$ Now, we differentiate with respect to \(u\): $$\frac{d v}{d u} = \frac{1}{u}.$$ Then, we rewrite the integral in terms of \(v\): $$\int_{\ln 2}^{\ln 3} \frac{1}{\ln^2 v}\,\mathrm{d}v.$$ Now, we have a much simpler integral to evaluate: $$\int_{\ln 2}^{\ln 3} \frac{1}{\ln^2 v}\,\mathrm{d}v.$$ Integrating, we find that the antiderivative is \(-\frac{1}{\ln v}\), and so our definite integral evaluates to $$\left[-\frac{1}{\ln v}\right]_{\ln 2}^{\ln 3} = \frac{1}{\ln 2} - \frac{1}{\ln 3}.$$