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Chapter 7: Problem 29

Use a table of integrals to determine the following indefinite integrals. These integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table. $$\int \frac{d x}{\sqrt{x^{2}-6 x}}, x>6$$

Short Answer

Expert verified
Question: Evaluate the indefinite integral: $$\int \frac{dx}{\sqrt{x^2 - 6x}}, x>6$$ Answer: $$\int \frac{dx}{\sqrt{x^2 - 6x}} =\cosh^{-1}\left(\frac{x-3}{3}\right) + C$$

Step by step solution

01

Complete the square

The first step is to complete the square for the denominator: \(x^2 - 6x = (x - 3)^2 - 9\) So our integral becomes: $$\int \frac{dx}{\sqrt{(x - 3)^2 - 9}}$$
02

Perform substitution

Let's use substitution method. Let \(x - 3 = 3\sinh{u}\), so \(dx = 3\cosh{u} \, du\). The integral now becomes: $$\int \frac{3\cosh{u}}{\sqrt{9\sinh^2{u}}} du$$ Simplify the integral: $$\int \frac{\cosh{u}}{\sqrt{\sinh^2{u}}} du$$
03

Use table of integrals

Now, our integral appears in a familiar form found in the table of integrals: $$\int \frac{\cosh{u}}{\sqrt{\sinh^2{u}}} du = \int \operatorname{csch}(u)\cosh(u) \, du$$ Using the table of integrals, we know that: $$\int \operatorname{csch}(u)\cosh(u) \, du = \cosh^{-1}(u) + C$$ So our integral expression becomes: $$\cosh^{-1}(u) + C$$
04

Substitute back the original variable

Now substitute back \(x-3 = 3\sinh{u}\): $$\cosh^{-1}\left(\frac{x-3}{3}\right) + C$$ So, the indefinite integral is: $$\int \frac{dx}{\sqrt{x^2 - 6x}} = \cosh^{-1}\left(\frac{x-3}{3}\right) + C$$

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