Q. 40 - 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. 

$\int \frac{-3}{1+16x^2}dx.$

Step-by-Step Solution

Answer

The solution of the integral is $-\frac{3}{4}{\tan }^{-1}\left(4x\right)+C.$

Step 1. Given Information.

The given integral is$\int \frac{-3}{1+16x^2}dx.$

Step 2. Solve.

By solving the integral we get, 

$$\begin{gathered} \int \frac{-3}{1+16x^2}dx \\ =-3\int \frac{1}{1+16x^2}dx \\ =-3\int \frac{1}{1+{\left(4x\right)}^2}dx \\ =-\frac{3}{4}{\tan }^{-1}\left(4x\right)+C \end{gathered}$$

Step 3. Verification.

To verify the answer we differentiate $-\frac{3}{4}{\tan }^{-1}\left(4x\right)+C$ it.

On differentiating we get,

$$\begin{gathered} -\frac{3}{4}{\tan }^{-1}\left(4x\right)+C \\ =\frac{d}{dx}\left(-\frac{3}{4}{\tan }^{-1}\left(4x\right)\right)+\frac{d}{dx}C \\ =-\frac{3}{4}\frac{d}{dx}{\tan }^{-1}\left(4x\right)\times \frac{d}{dx}\left(4x\right)+\frac{d}{dx}C \\ =-\frac{3}{4}\times 4\left(\frac{1}{1+{\left(4x\right)}^2}\right)+0 \\ =\frac{-3}{1+16x^2} \end{gathered}$$

Hence proved.