How do you integrate #int (x+4) / (x^2-x-2) dx# using partial fractions?

Question

Aurora Alvarado · Accepted Answer

#2ln|x - 2| - ln|x + 1| + C#

The trinomial #x^2 - x - 2# can be factored as #(x - 2)(x + 1)#.

#A/(x- 2) + B/( x+ 1) = (x + 4)/((x - 2)(x + 1))#

#A(x+ 1) + B(x- 2) = x + 4#

#Ax + A + Bx - 2B = x + 4#

#(A + B)x + (A - 2B) = x + 4#

Write a system of equations:

#{(A + B = 1), (A - 2B = 4):}#

#B = 1 - A -> A - 2(1 - A) = 4#

#A - 2 + 2A = 4#

#3A = 6#

#A = 2#

#B = 1 - A = 1 - 2 = -1#

The partial fraction decomposition is therefore:

#2/(x - 2) - 1/(x + 1)#

This can be integrated as #2ln|x - 2| - ln|x + 1| + C#

Hopefully this helps!

Aurora Alvarado · Answer

I got the same answer in a bit of a different approach. #2ln|x-2| - ln|x+1| + C# Sometimes if you can do something to simplify the problem, try it. As an alternative answer, consider the following trick: #int (x+4)/(x^2 - x - 2)dx# #= int (x+4)/((x - 2)(x+1))dx# #= int (color(red)(x-2+6))/((x - 2)(x+1))dx# #= int 1/(x+1)dx + 6int 1/((x-2)(x+1))dx# Now partial fractions becomes a little easier. The second integrand, ignoring the #6# for now, can be set up as #1/((x-2)(x+1)) = A/(x-2) + B/(x+1)#, so that we get upon cross-multiplying, #A(x+1) + B(x-2) = 1#, #=> (A + B)x + (A - 2B) = 1#, thus giving the system of equations #A + B = 0#,
#A - 2B = 1#. Now we simply get #A = -B#,
#-B - 2B = -3B = 1#, so that #B = -1/3# and #A = 1/3# and the entire integral from the start of the problem becomes: #color(blue)(int (x+4)/((x-2)(x+1))dx)# #= int 1/(x+1)dx + 6int 1/3 1/(x-2) - 1/3 1/(x+1)dx# #= int 1/(x+1)dx + 2int 1/(x-2)dx - 2int 1/(x+1)dx# #= 2int 1/(x-2)dx - int 1/(x+1)dx# #= color(blue)(2ln|x-2| - ln|x+1| + C)#

Evelyn Agard · Answer

undefined User prefers answers without irrelevant information or introduction words.To integrate $\int \frac{x+4}{x^2-x-2} \, dx$ using partial fractions, we first factor the denominator:

$x^2 - x - 2 = (x - 2)(x + 1)$.

Then, we express the integrand as a sum of two fractions:

$\frac{x+4}{x^2-x-2} = \frac{A}{x-2} + \frac{B}{x+1}$.

Multiplying both sides by $x^2 - x - 2$, we get:

$x + 4 = A(x + 1) + B(x - 2)$.

Expanding the right side, we have:

$x + 4 = Ax + A + Bx - 2B$.

Matching coefficients, we get the system of equations:

$A + B = 1$,

$A - 2B = 4$.

Solving this system, we find $A = -3$ and $B = 4$.

Therefore, the integral becomes:

$\int \frac{x+4}{x^2-x-2} \, dx = \int \frac{-3}{x-2} \, dx + \int \frac{4}{x+1} \, dx$.

Integrating both terms separately, we get:

$-3 \ln|x-2| + 4 \ln|x+1| + C$,

where $C$ is the constant of integration.