Q. 38 - 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. 

$\int \frac{x^{-1}}{1-x^{-1}}dx$

Step-by-Step Solution

Answer

The solution of the integral is  $\ln \left|x-1\right|+C.$

Step 1. Given Information.

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

Step 2. Solve.

By solving the integral we get, 

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

Step 3. Verification.

To verify the answer we differentiate $\ln \left|x-1\right|+C$ it.

On differentiating we get,

$$\begin{gathered} \ln \left|x-1\right|+C \\ =\frac{d}{dx}\ln \left|x-1\right|+\frac{d}{dx}C \\ =\frac{1}{x-1}+0 \\ =\frac{1}{x-1} \\ =\frac{x^{-1}}{1-x^{-1}} \end{gathered}$$

Hence proved.