Question

How do you use partial fraction decomposition to decompose the fraction to integrate$\dfrac{7}{{{x^2} + 13x + 40}}$?

Answer 1

Partial fraction is the decomposition of an equation which is of the form$\dfrac{{A(x)}}{{B(x)}}$. This method is only used when the fraction is linear in the denominator part. If the equation is raised to some power then this method fails.

Complete step by step answer:
We should first convert the equation to a linear form.
$= \dfrac{7}{{{x^2} + 13x + 40}} \\\ = \dfrac{7}{{{x^2} + 5x + 8x + 40}} \\\ = \dfrac{7}{{x(x + 5) + 8(x + 5)}} \\\ = \dfrac{7}{{(x + 8)(x + 5)}} \\\$
Now , we use partial fractions,
$\dfrac{A}{{x + 8}} + \dfrac{B}{{x + 5}} = \dfrac{7}{{(x + 8)(x + 5)}} \\\ A(x + 5) + B(x + 8) = 7 \\\ (A + B)x + 5A + 8B = 7 \\\$
Comparing the coefficients on both sides,
$A + B = 0 \\\ 5A + 8B = 7 \\\$
Solving the above equations simultaneously
$A = - B \\\ 5A + 8B = 7 \\\ 5( - B) + 8B = 7 \\\ 8B - 5B = 7 \\\ 3B = 7 \\\ B = \dfrac{7}{3},A = - \dfrac{7}{3} \\\$
Substituting these values to the main equation,
$= \int { - \dfrac{7}{3}} (\dfrac{1}{{x + 8}}) + \dfrac{7}{3}(\dfrac{1}{{x + 5}})dx \\\ = \dfrac{7}{3}\int { - \dfrac{1}{{x + 8}}} + \dfrac{1}{{x + 5}}dx \\\ = \dfrac{7}{3}[ - \log (x + 8) + \log (x + 5)] + C \\\ = \dfrac{7}{3}\log (\dfrac{{x + 5}}{{x + 8}}) + C \\\$
{Since,$\int {\dfrac{1}{{x + a}}} dx = \log (x + a) + C$}

Note: If $\dfrac{{A(x)}}{{B(x)}}$ is the equation then the partial fraction is applicable , if and only if, the degree of $A(x)$ is less than that of $B(x)$. If the value is greater, then this method fails and we need to look for some other alternative.