How do you integrate #(7x + 44)/(x^2 + 10x + 24)# using partial fractions?

Question

Isabella Ackerman · Accepted Answer

#= - ln (x+6) + 8 ln (x+4) + C#

#(7x + 44)/(x^2 + 10x + 24)#

#= (7x + 44)/( (x+6)(x+4) ) qquad qquad star#

#= A/(x+6) + B/(x+4) #

#= (A(x+4) + B(x+6) ) /((x+6) (x+4)) qquad qquad square#

so comparing #star# to #square#

x = -4 7(-4) + 44 = B(-4+6), B = 8

x = -6 7(-6) + 44 = A(-6+4), A = -1

so we have

#int dx qquad -1/(x+6) + 8/(x+4)#

#= - ln (x+6) + 8 ln (x+4) + C#

Charles Arocha · Answer

undefined To integrate $ \frac{7x + 44}{x^2 + 10x + 24} $ using partial fractions, you first factor the denominator $x^2 + 10x + 24$, which factors as $(x + 6)(x + 4)$. Then, you decompose the rational function into partial fractions. Since the denominator factors into distinct linear factors, the partial fraction decomposition will be of the form $ \frac{A}{x + 6} + \frac{B}{x + 4} $.

Now, you solve for $A$ and $B$ by equating coefficients:

$7x + 44 = A(x + 4) + B(x + 6)$

Expanding and collecting like terms:

$7x + 44 = (A + B)x + (4A + 6B)$

By comparing coefficients of like terms:

$A + B = 7$ (coefficients of $x$)

$4A + 6B = 44$ (constant terms)

Solving this system of linear equations gives $A = 2$ and $B = 5$.

Thus, the partial fraction decomposition of $ \frac{7x + 44}{x^2 + 10x + 24} $ is $ \frac{2}{x + 4} + \frac{5}{x + 6} $.

You can now integrate each term separately:

$$
\int \frac{7x + 44}{x^2 + 10x + 24} \, dx = \int \left(\frac{2}{x + 4} + \frac{5}{x + 6}\right) \, dx
$$

$$
= 2\ln|x + 4| + 5\ln|x + 6| + C
$$

where $C$ is the constant of integration.