How do you integrate #int(x+1)/((x-9)(x+3)(x-2))# using partial fractions?

Question

Ezra Adams · Accepted Answer

Set up a partial fraction decomposition and choose #x#-values to find the numerators of the partial fractions to get #int(x+1)/((x-9)(x+3)(x-2))dx=5/42lnabs(x-9)-1/30lnabs(x+3)-3/35lnabs(x-2)+C#.

Since all the factors in the denominator of #(x+1)/((x-9)(x+3)(x-2))# are linear (yay!), our partial fraction decomposition will be: #A/(x-9)+B/(x+3)+C/(x-2)# For now, let's ignore the integral and focus on decomposing this fraction: #(x+1)/((x-9)(x+3)(x-2))=A/(x-9)+B/(x+3)+C/(x-2)# #x+1=A(x+3)(x-2)+B(x-9)(x-2)+C(x-9)(x+3)# Let #x=9# to find the value of #A#: #9+1=A(9+3)(9-2)+B(9-9)(9-2)+C(9-9)(9+3)# #10=84A# #A=10/84=5/42# Let #x=-3# to find the value of #B#: #-3+1=A(-3+3)(-3-2)+B(-3-9)(-3-2)+C(-3-9)(-3+3)# #-2=60B# #B=-2/60=-1/30# And let #x=2# to find the value of #C#: #2+1=A(2+3)(2-2)+B(2-9)(2-2)+C(2-9)(2+3)# #3=-35C# #C=-3/35# Therefore, #(x+1)/((x-9)(x+3)(x-2))=(5/42)/(x-9)-(1/30)/(x+3)-(3/35)/(x-2)#. Our integral is now: #int(5/42)/(x-9)-(1/30)/(x+3)-(3/35)/(x-2)dx# Using the sum rule, this becomes: #int(5/42)/(x-9)dx-int(1/30)/(x+3)dx-int(3/35)/(x-2)dx# Integrating these gives: #5/42int1/(x-9)dx-1/30int1/(x+3)dx-3/35int1/(x-2)dx# #5/42lnabs(x-9)-1/30lnabs(x+3)-3/35lnabs(x-2)+C#

Ezra Adams · Answer

undefined To integrate $\frac{{x+1}}{{(x-9)(x+3)(x-2)}}$ using partial fractions, we first perform partial fraction decomposition. We express the given rational function as the sum of simpler fractions:

$\frac{{x+1}}{{(x-9)(x+3)(x-2)}} = \frac{{A}}{{x-9}} + \frac{{B}}{{x+3}} + \frac{{C}}{{x-2}}$

Next, we clear the denominators by multiplying both sides by $(x-9)(x+3)(x-2)$:

$x+1 = A(x+3)(x-2) + B(x-9)(x-2) + C(x-9)(x+3)$

We then expand and equate coefficients of like terms:

For $x^2$ term:
$1 = A + B + C$

For $x$ term:
$0 = -2A - 2B + 3C$

For constant term:
$0 = -6A - 18B - 9C$

Solving this system of equations gives us the values of $A$, $B$, and $C$. Subsequently, we integrate each term separately:

$\int \frac{{A}}{{x-9}} + \frac{{B}}{{x+3}} + \frac{{C}}{{x-2}} dx$

$= A\ln|x-9| + B\ln|x+3| + C\ln|x-2| + D$

Where $D$ is the constant of integration. Finally, we substitute the values of $A$, $B$, and $C$ obtained from the partial fraction decomposition to find the definite integral.