Chapter 5.7, Problem 22E

To determine

To calculate: The value of given integral

Answer to Problem 22E

The value of the given integral is $\ln \left(\frac{3}{8}\right)$

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_0^1\frac{x-4}{x^2-5x+6}}$

Calculation:

The denominator $x^2-5x+6$ is factored to $\left(x-2\right)\left(x-3\right)$

Now, do the partial function expansion of the integrand

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

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

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

The coefficients of $x$ must be equal and the constant terms are also equal. So

  $A+B=1\Rightarrow B=1-A\text{ and }-3A-2B=-1$

Solve this system of linear equation for $A\text{ and }B$ , by use of simple substitution

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

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

Rewrite the integral using the values found for $A\text{ and }B$ in the previous step

  ${\displaystyle {\int }_0^1\frac{2}{x-2}}-\frac{1}{x-3}$

These fractions can be integrated directly, no substitutions are necessary

  ${\left[2\ln \left|x-2\right|-\ln \left|x-3\right|\right]}_0^1$

Use the properties of logarithm to simplify the expression

  $\begin{array}{l}2\ln \left|1-2\right|-\ln \left|1-3\right|-2\ln \left|0-2\right|+\ln \left|0-3\right|\\ =2\ln \left(1\right)-\ln \left(2\right)-2\ln \left(2\right)+\ln \left(3\right)\\ =\ln \left(2^{-1}\right)+\ln \left(2^{-2}\right)+\ln \left(3\right)\\ =\ln \left(\frac{1}{2}\right)+\ln \left(\frac{1}{4}\right)+\ln \left(3\right)\\ =\ln \left(\frac{3}{8}\right)\end{array}$

Therefore, the value of the given integral is $\ln \left(\frac{3}{8}\right)$