Chapter 7, Problem 9RE

To determine

To evaluate the integral.

Answer to Problem 9RE

The required solution is , ${\displaystyle {\int }_0^1\frac{x}{x^2+5x+6}}dx=\text{ ln}\frac{256}{243}$ .

Explanation of Solution

Given information: The equation is, ${\displaystyle {\int }_0^1\frac{x}{x^2+5x+6}}dx$ .

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,

  ${\displaystyle {\int }_0^1\frac{x}{x^2+5x+6}}dx$

Use fraction decomposition so factor the denominator.

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

Substitute in $x=-2$ so you can solve A.

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

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

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

Substitute the value of A and B.

  $\begin{array}{l}\frac{x}{\left(x+2\right)\left(x+3\right)}=\frac{-2}{x+2}+\frac{3}{x+3}\\ {\displaystyle {\int }_0^1\left(\frac{-2}{x+2}\text{+}\frac{3}{x+3}\right)}dx\\ \text{ = }{\left[-2\ln \left|x+2\right|+3\ln \left|x+3\right|\right]}_0^1\\ \text{ = }{\left[\ln {\left(x+3\right)}^3-\ln {\left(x+2\right)}^2\right]}_0^1\\ \text{ = }{\left[\text{ln}\frac{{\left(x+2\right)}^3}{{\left(x+3\right)}^2}\right]}_0^1\\ \text{ = }{\left[\text{ln}\frac{{\left(x+3\right)}^3}{{\left(x+2\right)}^2}\right]}_0^1\\ \text{ = ln}\frac{{\left(1+3\right)}^3}{{\left(1+2\right)}^2}-\text{ln}\frac{{\left(0+3\right)}^3}{{\left(0+2\right)}^2}\\ \text{ = ln}\frac{{\left(4\right)}^3}{{\left(3\right)}^2}-\text{ln}\frac{{\left(3\right)}^3}{{\left(2\right)}^2}\\ \text{ = ln}\frac{64}{9}-\text{ln}\frac{27}{4}\\ \text{ = ln}\frac{64/9}{27/4}\\ \text{ = ln}\frac{256}{243}\end{array}$

Hence, the required solution is ${\displaystyle {\int }_0^1\frac{x}{x^2+5x+6}}dx=\text{ ln}\frac{256}{243}$ .