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Chapter 6: Problem 15

Find the integral. $$ \int \frac{x}{\sqrt{x^{2}+9}} d x $$

Short Answer

Expert verified
The integral of \( \int \frac{x}{\sqrt{x^{2}+9}} dx \) is \( \sqrt{x^{2} + 9} + C \) where \(C\) is the constant of integration.

Step by step solution

01

Perform substitution

We make the substitution \( u = x^2 + 9 \) and differentiate this expression to get \( du = 2xdx \). Now divide by 2 to keep the \( xdx \) term that appears in the integral, we get: \( \frac{1}{2}du = xdx \)
02

Substitute into the integral

Substitute into the integral \[ \int \frac{x}{\sqrt{x^{2}+9}} d x \] Now it becomes: \[ \frac{1}{2} \int \frac{du} {\sqrt{u}} \] as \(x dx = \frac{1}{2} du \) and \(\sqrt{x^{2}+9} = \sqrt{u}\).
03

Solve the integral

The new integral is easier to calculate. It is of form \( \int u^n du \) which has a standard result of \( \frac{u^{n+1}}{n+1} \). Applying it, we get: \[ \frac{1}{2} * 2\sqrt{u} = \sqrt{u} \] since the power \( n = -\frac{1}{2} \) and adding 1 to -\frac{1}{2}, we get \(\frac{1}{2}\). So it is \(\sqrt{u}\) after we integrate.
04

Substitute back

Now substitute back in to replace \(u\) with the original variable: \[ \sqrt{x^{2} + 9} \] We get the final result.

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