Chapter 7.5, Problem 5E

To determine

To evaluate the integral.

Answer to Problem 5E

The required solution is , ${\displaystyle \int \frac{x-12}{x^2-4}dx}=\ln \left(\frac{\left|x\right|}{{\left(x-4\right)}^2}\right)+C$ .

Explanation of Solution

Given information: The equation is, ${\displaystyle \int \frac{x-12}{x^2-4}dx}$ .

Concept used: Partial fraction decomposition.

If $f\left(x\right)=\frac{P\left(x\right)}{Q\left(x\right)}$ , where P and Q are polynomials with the degree of P less than the degree of Q,

and if $Q\left(x\right)$ can be written as a product of distinct linear factors, then $f\left(x\right)$ can be written as

a sum of rational functions with distinct linear denominators.

Calculation : The required solution is obtained as,

  ${\displaystyle \int \frac{x-12}{x^2-4x}dx}$

Use fraction decomposition so factor the denominator.

  $\begin{array}{l}\frac{x-12}{x^2-4x}=\frac{A}{x}+\frac{B}{x-4}\\ \frac{x-12}{x^2-4x}=\frac{A\left(x-4\right)+B\left(x\right)}{x\left(x-4\right)}\\ \frac{x-12}{x^2-4x}=\frac{A\left(x-4\right)+B\left(x\right)}{x^2-4x}\\ x-12=A\left(x-4\right)+B\left(x\right)\left(\text{cancel denominator both side}\right)\end{array}$

Substitute in $x=4$ so you can solve B.

  $\begin{array}{l}4-12=A\left(4-4\right)+B\left(4\right)\\ -8=A\left(0\right)+B\left(4\right)\\ B=-2\end{array}$

Substitute in $x=0$ so you can solve A.

  $\begin{array}{l}0-12=A\left(0-4\right)+B\left(0\right)\\ -12=A\left(-4\right)\\ A=3\end{array}$

Substitute the value of A and B.

  $\begin{array}{l}{\displaystyle \int \frac{x-12}{x^2-4}dx}={\displaystyle \int \left(\frac{3}{x}-\frac{2}{x-4}\right)}dx\\ =3\ln \left|x\right|-2\ln \left|x-4\right|+C\\ =\ln {\left|x\right|}^3-\ln {\left|x-4\right|}^2+C\\ =\ln \left(\frac{\left|x\right|}{{\left(x-4\right)}^2}\right)+C\end{array}$

Hence, the required solution is ${\displaystyle \int \frac{x-12}{x^2-4}dx}=\ln \left(\frac{\left|x\right|}{{\left(x-4\right)}^2}\right)+C$ .