Q. 62 - 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. (Hint for Exercise 54: $\tan \left(x\right) = \frac{\sin \left(x\right)}{\cos \left(x\right)}$).

$\int \frac{1}{(1+x)(1-x)}dx.$

Step-by-Step Solution

Answer

The value of the given integral is $\frac{1}{2}\ln \left(\frac{\left(1+x\right)}{\left(1-x\right)}\right)+c.$

Step 1. Given Information.

Given is a integral is $\int \frac{1}{(1+x)(1-x)}dx.$

Step 2. Formula involved.

$\int \frac{1}{x}dx = \ln \left(x\right)+c.$

Step 3. Formula involved.

$$\begin{gathered} \int \frac{1}{(1+x)(1-x)}dx \\ =\frac{1}{2}\int \frac{2}{(1+x)(1-x)}dx \\ = \frac{1}{2}\int \frac{(1+x)+(1-x)}{(1+x)(1-x)}dx \\ =\frac{1}{2}\left(\int \frac{1}{(1-x)}dx+ \int \frac{1}{(1+x)}dx\right) \\ =\frac{1}{2}\left(\frac{\ln (1-x)}{-1}+ \ln (1+x)\right) +c \\ =\frac{1}{2}\ln \left(\frac{\left(1+x\right)}{\left(1-x\right)}\right)+c. \end{gathered}$$