Chapter 5.7, Problem 23E

To determine

To calculate: The value of given integral

Answer to Problem 23E

The value of the given integral is $\frac{1}{2}\ln \left(\frac{3}{2}\right)$

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_2^3\frac{1}{x^2-1}}$

Calculation:

The denominator $x^2-1$ is factored to $\left(x+1\right)\left(x-1\right)$

Now, do the partial function expansion of the integrand

  $\frac{1}{\left(x+1\right)\left(x-1\right)}=\frac{A}{x+1}+\frac{B}{x-1}$

Here, $A\text{ and }B$ are constants. To find the values of $A\text{ and }B$ multiply both sides of the equation by $\left(x+1\right)\left(x-1\right)$ . This results to two unknowns

  $\begin{array}{l}1=A\left(x-1\right)+B\left(x+1\right)\\ 1=Ax-A+Bx+B\\ 1=x\left(A+B\right)+\left(-A+B\right)\end{array}$

The coefficients of $x$ must be equal and the constant terms are also equal. So

  $A+B=0\text{ and }1=B-A\Rightarrow B=1+A$

Solve this system of linear equation for $A\text{ and }B$ , by use of simple substitution and algebra

  $\begin{array}{l}A+\left(1+A\right)0\Rightarrow 2A=-1\Rightarrow A=-\frac{1}{2}\\ B=1+A=1-\frac{1}{2}=\frac{1}{2}\end{array}$

So, $A=-\frac{1}{2}\text{ and }B=\frac{1}{2}$

Rewrite the integral in expanded form. Factor out $\frac{1}{2}$ in order to simplify the expression

  $\frac{1}{2}{\displaystyle {\int }_2^3\left[\frac{1}{x-1}-\frac{1}{x+1}\right]}dx$

Integrate by the use of the identity ${\displaystyle \int \frac{1}{u}=\ln \left|u\right|+C}$ .These fractions can be integrated directly, no substitutions are necessary

  $\frac{1}{2}{\left[\ln \left|x-1\right|-\ln \left|x+1\right|\right]}_2^3$

Use the properties of logarithm to simplify the expression

  $\begin{array}{l}\frac{1}{2}\left[\ln \left|3-1\right|-\ln \left|3+1\right|-\ln \left|2-1\right|+\ln \left|2+1\right|\right]\\ =\frac{1}{2}\left[\ln \left(2\right)-\ln \left(4\right)-\ln \left(1\right)+\ln \left(3\right)\right]\\ =\frac{1}{2}\ln \left(\frac{3}{2}\right)\end{array}$

Therefore, the value of the given integral is $\frac{1}{2}\ln \left(\frac{3}{2}\right)$