Question

Solve $\int \frac{1}{x^2+9}dx$ the following two ways:

(a) with the trigonometric substitution x = 3 tan u;

(b) with algebra and the derivative of the arctangent.

Short answer

Part (a) The solution of the integral is $\frac{1}{3}{\tan }^{-1}\frac{x}{3}+C.$

Part (b) The solution of the integral is $\frac{1}{3}{\tan }^{-1}\frac{x}{3}+C.$

Step-by-step solution

Part (a) Step 1. Given Information.

The given integral is$\int \frac{1}{x^2+9}dx.$

Part (a) Step 2. Solve. 

We have to solve the integral with the substitution $x=3\tan u,$so the derivative is$dx=3se{c}^{2}udu.$

Let's solve the integral by substituting x,

$$\begin{gathered} \int \frac{1}{x^2+9}dx \\ =\int \frac{1}{{\left(3\tan u\right)}^2+9}\left(3se{c}^{2}u\right)du \\ =\int \frac{3se{c}^{2}u}{9{\tan }^{2}u+9}du \\ =\int \frac{3se{c}^{2}u}{9\left({\tan }^{2}u+1\right)}du \\ =\frac{1}{3}\int \frac{se{c}^{2}u}{{\tan }^{2}u+1}du \\ \mathrm{Usethe}\mathrm{identity}1+ta{n}^{2}x=se{c}^{2}x \\ =\frac{1}{3}\int \frac{se{c}^{2}u}{se{c}^{2}u}du \\ =\frac{1}{3}\int 1du \\ =\frac{1}{3}u+C \\ \mathrm{Substitute}\mathrm{back}u, \\ =\frac{1}{3}{\tan }^{-1}\left(\frac{x}{3}\right)+C \end{gathered}$$

Part (b) Step 1. Solve. 

We have to solve the integral with algebra and the derivative of the arctangent.

As we know the derivative of the arctangent is $\frac{1}{x^2+1}.$

Now, let's solve the integral,

$$\begin{gathered} \int \frac{1}{x^2+9}dx \\ =\int \frac{1}{9\left({\displaystyle \frac{x^2}{9}}+1\right)}dx \\ =\frac{1}{9}\int \frac{1}{\left({\left(\frac{x}{3}\right)}^2+1\right)}dx \\ \mathrm{Use}\mathrm{the}\mathrm{derivative}\mathrm{of}\mathrm{arctangent} \\ =\frac{1}{9}\times 3{\tan }^{-1}\frac{x}{3}+C \\ =\frac{1}{3}{\tan }^{-1}\frac{x}{3}+C \end{gathered}$$