Q. 47 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

Step 1. Given Information

Solving the given integrals. 

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

Step 2. Solving the given integral using substitution method.

Let

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

Step 3. This substitution changes the integral into

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