Q. 42 - Calculus

Question

Use integration formulas to solve each integral in Exercises 21–62. You may have to use algebra, educated guess-and-check, and/or recognize an integrand as the result of a product, quotient, or chain rule calculation. Check each of your answers by differentiating.  

$\int \frac{1}{4+x^2}dx$

Step-by-Step Solution

Answer

The solution of the integral is $\frac{1}{2} {\tan }^{-1}\left(\frac{x}{2}\right)+C.$

Step 1. Given Information.

The given integral is $\int \frac{1}{4+x^2}dx.$

Step 2. Solve.

By solving the integral we get, 

$$\begin{gathered} \int \frac{1}{4+x^2}dx \\ =\int \frac{1}{4\left(1+{\displaystyle \frac{x^2}{4}}\right)}dx \\ =\frac{1}{4}\int \frac{1}{1+{\left({\displaystyle \frac{x}{2}}\right)}^2}dx \\ =\frac{1}{4}\left(2\right) {\tan }^{-1}\left(\frac{x}{2}\right)+C \\ =\frac{1}{2} {\tan }^{-1}\left(\frac{x}{2}\right)+C \end{gathered}$$

Step 3. Verification.

To verify the answer we differentiate $\frac{1}{2} {\tan }^{-1}\left(\frac{x}{2}\right)+C$ it.

On differentiating we get,

$$\begin{gathered} \frac{1}{2} {\tan }^{-1}\left(\frac{x}{2}\right)+C \\ =\frac{d}{dx}\frac{1}{2} {\tan }^{-1}\left(\frac{x}{2}\right)+\frac{d}{dx}C \\ =\frac{1}{2}\frac{d}{dx}{\tan }^{-1}\left(\frac{x}{2}\right)+\frac{d}{dx}C \\ =\frac{1}{2}\times \frac{1}{2}\left(\frac{1}{1+{\left({\displaystyle \frac{x}{2}}\right)}^2}\right)+0 \\ =\frac{1}{4+x^2} \end{gathered}$$

Hence proved.