Q. 41 - 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.  

$\int \frac{1}{\left|3x\right|\sqrt{9x^2-1}}dx$

Step-by-Step Solution

Answer

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

Step 1. Given Information.

The given integral is $\int \frac{1}{\left|3x\right|\sqrt{9x^2-1}}dx.$

Step 2. Solve.

By solving the integral we get, 

$$\begin{gathered} \int \frac{1}{\left|3x\right|\sqrt{9x^2-1}}dx \\ =\int \frac{1}{\left|3x\right|\sqrt{{\left(3x\right)}^2-1}}dx \\ \mathrm{Use} \left[\int \frac{1}{\left|\mathrm{x}\right|\sqrt{{\left(\mathrm{x}\right)}^2-1}}\mathrm{dx} ={\sec }^{-1}\left(\mathrm{x}\right)+\mathrm{C}\right] \\ =\frac{1}{3}se{c}^{-1}\left(3x\right)+C \end{gathered}$$

Step 3. Verification.

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

On differentiating we get,

$$\begin{gathered} \frac{1}{3}se{c}^{-1}\left(3x\right)+C \\ =\frac{d}{dx}\frac{1}{3}se{c}^{-1}\left(3x\right)+\frac{d}{dx}C \\ =\frac{1}{3}\frac{d}{dx}se{c}^{-1}\left(3x\right)+\frac{d}{dx}C \\ =\frac{1}{3}\times 3\left(\frac{1}{\left|3x\right|\sqrt{{\left(3x\right)}^2-1}}\right)+0 \\ =\frac{1}{\left|3x\right|\sqrt{9x^2-1}} \end{gathered}$$

Hence proved.