How do you integrate #(1x^(2)+4x+12)/((x+2)(x^(2)+4)) # using partial fractions?

1 Answer
Aug 23, 2017

Write the partial fraction equation:

#(x^2+4x+12)/((x+2)(x^2+4)) = A/(x+2) + (Bx)/(x^2+4) + C/(x^2+1)#

Multiply both sides by #(x+2)(x^2+4)#:

#x^2+4x+12 = A(x^2+4) + (Bx)(x+2) + C(x+2)" [1]"#

Make B and C disappear by letting #x = -2#

#(-2)^2+4(-2)+12 = A((-2)^2+4)#

#8 = A(8)#

#A = 1#

Substitute the value for A into equation [1]:

#x^2+4x+12 = (x^2+4) + (Bx)(x+2) + C(x+2)" [2]"#

Make B disappear by letting #x = 0#

#0^2+4(0)+12 = (0^2+4) + C(0+2)#

#8 =2C#

#C = 4#

Substitute the value for C into equation [2]:

#x^2+4x+12 = (x^2+4) + (Bx)(x+2) + 4(x+2)" [3]"#

Let (x = 1):

#1^2+4(1)+12 = (1^2+4) + (B(1))((1)+2) + 4(1+2)#

#0 = 3B#

#B = 0#

Remove the term for B from equation [3]

#x^2+4x+12 = (x^2+4) + 4(x+2)#

Divide by #(x+2)(x^2+4)#:

#(x^2+4x+12)/((x+2)(x^2+4)) = 1/(x+2) + 4/(x^2+4)" [4]"#

Equation [4] gives us the template for the integrals:

#int(x^2+4x+12)/((x+2)(x^2+4))dx = int1/(x+2)dx + 4int1/(x^2+1)dx#

Both integrals are well known:

#int(x^2+4x+12)/((x+2)(x^2+4))dx = ln|x+2| + 4tan^-1(x)+ C#