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Chapter 7: Problem 7

Use a table of integrals to determine the following indefinite integrals. $$\int \frac{d x}{\sqrt{x^{2}+16}}$$

Short Answer

Expert verified
Question: Find the indefinite integral of the function $$\int \frac{dx}{\sqrt{x^2 + 16}}$$ using a table of integrals. Answer: The indefinite integral of the given function is $$\int \frac{dx}{\sqrt{x^2 + 16}} = \ln\left|x + \sqrt{x^2 + 16}\right| + C$$, where C is the constant of integration.

Step by step solution

01

Identify the integral from the table of integrals

We look for the general form of the integral in a table of integrals. The integral \(\int \frac{dx}{\sqrt{x^2 + a^2}}\) is found in the table of integrals. The corresponding formula is as follows: $$ \int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\left|x + \sqrt{x^2 + a^2}\right|+ C$$ where \(a\) is a constant and \(C\) is the constant of integration.
02

Apply the formula to the given integral

In our case, the given integral is: $$\int \frac{dx}{\sqrt{x^2 + 16}}$$ Here, we can see that \(a^2 = 16\), which gives us \(a = 4\). Now we can apply the formula from the table of integrals: $$ \int \frac{dx}{\sqrt{x^2 + 4^2}} = \ln\left|x + \sqrt{x^2 + 4^2}\right| + C$$
03

Simplify the integral

Now we can simplify the integral: $$ \int \frac{dx}{\sqrt{x^2 + 4^2}} = \ln\left|x + \sqrt{x^2 + 16}\right| + C$$ So, the indefinite integral of the given function is: $$\int \frac{dx}{\sqrt{x^2 + 16}} = \ln\left|x + \sqrt{x^2 + 16}\right| + C$$

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