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Chapter 7: Problem 8

Use a table of integrals to determine the following indefinite integrals. $$\int \frac{d x}{\sqrt{x^{2}-25}}$$

Short Answer

Expert verified
Question: Find the indefinite integral of the function \(\frac{1}{\sqrt{x^2 - 25}}\). Answer: The indefinite integral of the given function is \(\int \frac{dx}{\sqrt{x^2 - 25}} = \ln|x + \sqrt{x^2 - 25}| + C\).

Step by step solution

01

Identify the integral in the table of integrals

We can find the similar integral in a table of integrals that has the following form: $$\int \frac{dx}{\sqrt{x^2 - a^2}}$$ Where \(a\) is a constant. In our given integral, we have a similar form, and our \(a\) is equal to 5, as \(x^2 - 25 = x^2 - (5)^2\).
02

Apply the integral from the table of integrals

According to the table of integrals, we have the following standard result for our integral: $$\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln|x + \sqrt{x^2 - a^2}| + C$$ Now, we will apply this result to our integral with \(a = 5\):
03

Find the indefinite integral of the given function

Now, we will substitute the value of \(a\) in the result obtained above: $$\int \frac{dx}{\sqrt{x^2 - 5^2}} = \ln|x + \sqrt{x^2 - 5^2}| + C$$ So, the indefinite integral of the given function is: $$\int \frac{dx}{\sqrt{x^2 - 25}} = \ln|x + \sqrt{x^2 - 25}| + C$$

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