Chapter 7, Problem 23RE

To determine

To evaluate the integral.

Answer to Problem 23RE

The required solution is , ${\displaystyle \int \frac{25}{x^2-25}dx}=\frac{5}{2}\left(\ln \left|x-5\right|-\ln \left|x+5\right|\right)+C$ .

Explanation of Solution

Given information: The equation is, ${\displaystyle \int \frac{25}{x^2-25}dx}$ .

Concept used: : Indefinite integral.

The family of all antiderivatives of a function $f\left(x\right)$ is a indefinite integral of $f$

With respect to $x$ and is denoted by ${\displaystyle \int f\left(x\right).}$

If F is an function such that $F\text{'}\left(x\right)=f\left(x\right)$ , then ${\displaystyle \int f\left(x\right)}dx=F\left(x\right)+C,$

Where C is an arbitrary constant, called the constant of integration.

Calculation : The required solution is obtained as,

  ${\displaystyle \int \frac{25}{x^2-25}dx}$

Use fraction decomposition so factor the denominator.

  $\begin{array}{l}{\displaystyle \int \frac{25}{x^2-25}dx}={\displaystyle \int \frac{25}{\left(x+5\right)\left(x-5\right)}dx}\\ ={\displaystyle \int \frac{A\left(x-5\right)+B\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}dx}\\ =A\left(x-5\right)+B\left(x+5\right)\end{array}$

Substitute in $x=-5$ so you can solve A.

  $\begin{array}{l}\text{ 25}=A\left(-5-5\right)+B\left(-5+5\right)\\ \text{ 25}=-10A+B\left(0\right)\\ -\text{2}\text{.5}=A\end{array}$

Substitute in $x=5$ so you can solve B

  $\begin{array}{l}\text{ 25}=A\left(5-5\right)+B\left(5+5\right)\\ \text{ 25}=A\left(0\right)+B\left(10\right)\\ \text{ 2}\text{.5}=B\end{array}$

Substitute the value of A and B.

  $\begin{array}{l}{\displaystyle \int \frac{25}{x^2-25}dx}={\displaystyle \int \frac{2.5\left(x+5\right)-2.5\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}}dx\\ ={\displaystyle \int \frac{2.5}{\left(x-5\right)}-\frac{2.5}{x+5}}dx\\ =\frac{5}{2}{\displaystyle \int \frac{1}{\left(x-5\right)}-\frac{1}{x+5}}dx\\ =\frac{5}{2}{\displaystyle \int \frac{1}{\left(x-5\right)}dx-\frac{5}{2}{\displaystyle \int \frac{1}{x+5}}}dx\\ =\frac{5}{2}\left(\ln \left|x-5\right|-\ln \left|x+5\right|\right)+C\end{array}$

Hence, the required solution is ${\displaystyle \int \frac{25}{x^2-25}dx}=\frac{5}{2}\left(\ln \left|x-5\right|-\ln \left|x+5\right|\right)+C$ .