Chapter 7.5, Problem 6E

To determine

To evaluate the integral.

Answer to Problem 6E

The required solution is , ${\displaystyle \int \frac{2x+16}{x^2+x-6}dx}=-2\ln \left|x+3\right|4\ln \left|x-2\right|+C$ .

Explanation of Solution

Given information: The equation is, ${\displaystyle \int \frac{2x+16}{x^2+x-6}dx}$ .

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,

  ${\displaystyle \int \frac{2x+16}{x^2+x-6}}dx$

Use fraction decomposition so factor the denominator.

  $\begin{array}{l}\frac{2x+16}{x^2+x-6}=\frac{A}{x+3}+\frac{B}{x-2}\\ \frac{2x+16}{x^2+x-6}=\frac{A\left(x-2\right)+B\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\\ \frac{2x+16}{x^2+x-6}=\frac{A\left(x-2\right)+B\left(x+3\right)}{x^2+x-6}\\ 2x+16=A\left(x-2\right)+B\left(x+3\right)\left(\text{cancel denominator both side}\right)\end{array}$

Substitute in $x=2$ so you can solve B.

  $\begin{array}{l}2\left(2\right)+16=A\left(2-2\right)+B\left(2+3\right)\\ 20=A\left(0\right)+B\left(5\right)\\ 4=B\end{array}$

Substitute in $x=-3$ so you can solve A.

  $\begin{array}{l}2\left(-3\right)+16=A\left(-3-2\right)+B\left(-3+3\right)\\ 10=A\left(-5\right)+B\left(0\right)\\ -2=A\end{array}$

Substitute the value of A and B.

  $\begin{array}{l}{\displaystyle \int \frac{2x+16}{x^2+x-6}dx}={\displaystyle \int \frac{-2}{x+3}}dx+{\displaystyle \int \frac{4}{x-2}dx}\\ =-2{\displaystyle \int \frac{1}{x+3}}dx+4{\displaystyle \int \frac{1}{x-2}dx}\\ =-2\ln \left|x+3\right|4\ln \left|x-2\right|+C\end{array}$

Hence, the required solution is ${\displaystyle \int \frac{2x+16}{x^2+x-6}dx}=-2\ln \left|x+3\right|4\ln \left|x-2\right|+C$ .