Chapter 5.7, Problem 32E

To determine

To calculate: The value of given integral by using long division

Answer to Problem 32E

The value of the given integral is $\frac{3}{2}+\ln \left(\frac{3}{2}\right)$

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_0^1\frac{x^3-4x-10}{x^2-x-6}dx}$

Formula used:

  ${\displaystyle \int \frac{1}{x}dx=\ln \left|x\right|+C}$

Calculation:

Use long division to rewrite the original integrand

  ${\displaystyle {\int }_0^1\left(x+1+\frac{3x-4}{x^2-x-6}\right)}dx$

Factor the denominator as the product of the linear factors

  $x^2-x-6=\left(x+2\right)\left(x-3\right)$

Now the leftmost term of the given integral can be written as,

  $\frac{3x-4}{x^2-x-6}=\frac{3x-4}{\left(x+2\right)\left(x-3\right)}=\frac{A}{x+2}+\frac{B}{x-3}$

Here, $A\text{ and }B$ are the two constants. To find the value of $A\text{ and }B$ multiply both sides by $\left(x+2\right)\left(x-3\right)$

  $\begin{array}{l}3x-4=A\left(x-3\right)+B\left(x+2\right)\\ 3x-4=Ax-3A+Bx+2B\\ 3x-4=x\left(A+B\right)-3A+2B\end{array}$

The coefficient of $x$ must be equal and the constant terms are all equal. So

  $A+B=3\text{ and }-3A+2B=-4$

Solving this system of linear equations of $A\text{ and }B$ the value is found as,

  $\begin{array}{l}A+B=3\text{ and }-3A+2B=-4\\ A+B=3\Rightarrow B=3-A\\ 2B=-4+3A\Rightarrow 6-2A=-4+3A\Rightarrow A=2,B=1\end{array}$

So, $A=2\text{ and }B=1$

Now, the integral is formed the term,

  ${\displaystyle {\int }_0^1\left(x+1+\frac{2}{x+2}+\frac{1}{x-3}\right)dx}$

The obtained integral can be simplified as,

  $\begin{array}{l}{\displaystyle {\int }_0^1\left(x+1+\frac{2}{x+2}+\frac{1}{x-3}\right)dx}\\ ={\left[\frac{x^2}{2}+x+2\ln \left|x+2\right|+\ln \left|x-3\right|\right]}_0^1\\ =\left[\frac{1}{2}+1+\ln {\left(3\right)}^2+\ln \left(2\right)\right]-\left[\frac{0^2}{2}+0+\ln {\left(2\right)}^2+\ln \left(3\right)\right]\\ =\frac{3}{2}+\ln \left(18\right)-\ln \left(12\right)\\ =\frac{3}{2}+\ln \left(\frac{3}{2}\right)\end{array}$

Therefore, the value of the given integral is $\frac{3}{2}+\ln \left(\frac{3}{2}\right)$