Question

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

$\int \frac{1}{3x+1}\mathrm{d}x$

Short answer

The solution of the given integral is $\int \frac{1}{3x+1}\mathrm{d}x=\frac{1}{3}\ln \left(3\mathrm{x}+1\right)+\mathrm{C}$.

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

$\int \frac{1}{3x+1}\mathrm{d}x$

Step 2. Solving the given integral using substitution method. 

Let

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

Step 3. This substitution changes the integral into 

$$\begin{gathered} \int \frac{1}{3x+1}\mathrm{d}x=\frac{1}{3}\int \frac{1}{u}\mathrm{du} \\ \int \frac{1}{3\mathrm{x}+1}\mathrm{dx}=\frac{1}{3}\ln \left(\mathrm{u}\right)+\mathrm{C} \\ \int \frac{1}{3\mathrm{x}+1}\mathrm{dx}=\frac{1}{3}\ln \left(3\mathrm{x}+1\right)+\mathrm{C} \end{gathered}$$