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Chapter 6: Problem 15

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

Short Answer

Expert verified
Question: Evaluate the following integral: $$\int_{e}^{e^{2}} \frac{d x}{x \ln ^{3} x}$$ Answer: $$\frac{1}{4}$$

Step by step solution

01

Apply substitution

First, let's substitute \(u = \ln x\). This means \(x = e^u\). Now, we need to determine the differential \(du\) in terms of \(dx\). Differentiate both sides with respect to x: $$\frac{du}{dx} = \frac{1}{x}$$ or $$du = \frac{dx}{x}$$ Now let's update the integral accordingly: $$\int_{e}^{e^{2}} \frac{d x}{x \ln ^{3} x} = \int_{\ln e}^{\ln e^2} \frac{du}{u^3}$$ Observe that the limits of integration also changed accordingly, coming from the new substitution.
02

Evaluate the updated integral

Now, we can rewrite the integral as: $$\int_{1}^{2} u^{-3}du$$ We can now find the antiderivative: $$\int u^{-3}du = \frac{u^{-2}}{-2} + C$$
03

Apply the limits of integration

Apply the limits of integration to the antiderivative: $$\left[\frac{-1}{2u^{2}}\right]_{1}^{2} = \frac{-1}{8} - \frac{-1}{2}$$ $$= \frac{-1}{8} + \frac{1}{2} = \frac{1}{4}$$
04

Convert back to original variable

Now, recall that \(u = \ln x\). Thus, we have: $$\frac{1}{4}$$ There is no need to include absolute values, as the result is a constant. The evaluated integral is: $$\int_{e}^{e^{2}} \frac{d x}{x \ln ^{3} x} = \frac{1}{4}$$

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