Question

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

$\int \frac{1-x}{1+3x^2}\mathrm{d}x$

Short answer

The solution of the given integral is $\int \frac{1-x}{1+3x^2}\mathrm{d}x=\frac{1}{3}{\tan }^{-1}3\mathrm{x}-\frac{1}{6}\ln \left(1+3{\mathrm{x}}^2\right)+\mathrm{C}$.

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

$\int \frac{1-x}{1+3x^2}\mathrm{d}x$

Step 2. Solving the given integral using substitution method. 

Let

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

Step 3. This substitution changes the integral into 

$$\begin{gathered} \int \frac{1-x}{1+3x^2}\mathrm{d}x=\int \left(\frac{1}{1+3x^2}-\frac{x}{1+3x^2}\right)\mathrm{d}x \\ \int \frac{1-x}{1+3x^2}\mathrm{d}x=\int \frac{1}{1+3x^2}\mathrm{d}x-\int \frac{x}{1+3x^2}\mathrm{d}x \\ \int \frac{1-x}{1+3x^2}\mathrm{d}x=\frac{1}{3}\int \frac{1}{1+v^2}\mathrm{dv}-\frac{1}{6}\int \frac{1}{u}\mathrm{du} \\ \int \frac{1-\mathrm{x}}{1+3{\mathrm{x}}^2}\mathrm{dx}=\frac{1}{3}{\tan }^{-1}\mathrm{v}-\frac{1}{6}\ln \left(\mathrm{u}\right)+\mathrm{C} \\ \int \frac{1-\mathrm{x}}{1+3{\mathrm{x}}^2}\mathrm{dx}=\frac{1}{3}{\tan }^{-1}3\mathrm{x}-\frac{1}{6}\ln \left(1+3{\mathrm{x}}^2\right)+\mathrm{C} \end{gathered}$$