Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

$\int \frac{x}{\sqrt{3x^2+1}}dx$

Short answer

The solution of the given integral is $\int \frac{x}{\sqrt{3x^2+1}}dx=\frac{1}{3}{(3x^2+1)}^{1/2}+C$.

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

$\int \frac{x}{\sqrt{3x^2+1}}dx$

Step 2. Solving the given integral using substitution method. 

Let

$$\begin{gathered} u=3x^2+1 \\ \frac{du}{dx}=6x \\ du=6xdx \\ \frac{1}{6}du=xdx \end{gathered}$$

Step 3. This substitution changes the integral into 

$$\begin{gathered} \int \frac{x}{\sqrt{3x^2+1}}dx=\frac{1}{6}\int \frac{1}{\sqrt{u}}du \\ \int \frac{x}{\sqrt{3x^2+1}}dx=\frac{1}{6}\int \frac{1}{u^{1/2}}du \\ \int \frac{x}{\sqrt{3x^2+1}}dx=\frac{1}{6}\int u^{-1/2}du \\ \int \frac{x}{\sqrt{3x^2+1}}dx=\frac{1}{6}\left[\frac{u^{-1/2+1}}{-1/2+1}\right]+C \\ \int \frac{x}{\sqrt{3x^2+1}}dx=\frac{1}{6}\left[\frac{u^{1/2}}{1/2}\right]+C \\ \int \frac{x}{\sqrt{3x^2+1}}dx=\frac{1}{6}·2·u^{1/2}+C \\ \int \frac{x}{\sqrt{3x^2+1}}dx=\frac{1}{3}{(3x^2+1)}^{1/2}+C \end{gathered}$$