Chapter 7, Problem 10RE

To determine

To evaluate the integral.

Answer to Problem 10RE

The required solution is , ${\displaystyle {\int }_1^2\frac{2x+6}{x^2-3x}}dx\text{ = }-6\ln 2$ .

Explanation of Solution

Given information: The equation is, ${\displaystyle {\int }_1^2\frac{2x+6}{x^2-3x}}dx$ .

Concept used: Definite integrals,

The definite integral of $f\left(x\right)$ over $\left[a,b\right]$ , denoted by ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx,}$ is given by, ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx,}={\left[F\left(\text{b}\right)\right]}_a^b=F\left(b\right)-F\left(a\right)$

Calculation : The required solution is obtained as,

  ${\displaystyle {\int }_1^2\frac{2x+6}{x^2-3x}}dx={\displaystyle {\int }_1^2\frac{2x+6}{x\left(x-3\right)}}dx$

Use fraction decomposition so factor the denominator.

  $\begin{array}{l}\frac{2x+6}{x\left(x-3\right)}=\frac{A}{x}+\frac{B}{x-3}\\ =\frac{A\left(x-3\right)}{x\left(x-3\right)}+\frac{Bx}{x\left(x-3\right)}\\ =\frac{A\left(x-3\right)+Bx}{\left(x+2\right)\left(x+3\right)}\\ \text{ 2}x+6=A\left(x-3\right)+Bx\end{array}$

Substitute in $x=3$ so you can solve B.

  $\begin{array}{l}\text{ 2}\left(3\right)+6=A\left(3-3\right)+B\left(3\right)\\ 12=A\left(0\right)+B\left(3\right)\\ \text{ 1}2=3B\\ B=4\end{array}$

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

  $\begin{array}{l}\text{ 2}\left(0\right)+6=A\left(0-3\right)+B\left(0\right)\\ 0+6=A\left(-3\right)\\ 6=-3A\\ -2=A\end{array}$

Substitute the value of A and B.

  $\begin{array}{l}{\displaystyle {\int }_1^2\frac{2x+6}{x^2-3x}}dx={\displaystyle {\int }_1^2\left(\frac{A}{x}+\frac{B}{x-3}\right)}dx\\ ={\displaystyle {\int }_1^2\left(\frac{-2}{x}+\frac{4}{x-3}\right)}dx\\ ={\displaystyle {\int }_1^2\left(\frac{4}{x-3}-\frac{2}{x}\right)}dx\\ \text{ = }{\left[\left(4\ln \left|x-3\right|-2\ln \left|x\right|\right)\right]}_1^2\\ \text{ = }\left(4\ln \left|2-3\right|-2\ln \left|2\right|\right)-\left(4\ln \left|1-3\right|-2\ln \left|1\right|\right)\\ \text{ = }\left(4\ln \left|-1\right|-2\ln 2\right)-\left(4\ln \left|-2\right|-2\left(0\right)\right)\\ \text{ = }4\ln 1-2\ln 2-4\ln 2\\ \text{ = }4\left(0\right)-6\ln 2\\ \text{ = }-6\ln 2\end{array}$

Hence, the required solution is ${\displaystyle {\int }_1^2\frac{2x+6}{x^2-3x}}dx\text{ = }-6\ln 2$ .