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

1 Answer
Apr 25, 2017

# int \ (2x+1) /( (x-2)(x^2+4)) \ dx =5/8 ln|x-2| - 5/16 ln|x^2+4| - 3/8arctan(x/2) + C#

Explanation:

Let:

# I = int \ (2x+1) /( (x-2)(x^2+4)) \ dx #

We can decompose the integrand into partial fraction; as:

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

Leading to:

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

Put #x=2 => 5=8A => A=5/8#

Comparing Coefficients:

# "Coeff"(x^2) => 0=A+B#
# "Coeff"(x^0) => 1 = 4A-2C#

And so:

# B=-5/8#
# 2C = 20/8-1 => C=3/4 #

Hence the partial fraction decomposition gives us:

# I = int \ (5/8)/(x-2) + ((-5/8)x+(6/8))/(x^2+4) \ dx #
# \ \ = 5/8 int \ 1/(x-2) \ dx - 1/8int \ (5x-6)/(x^2+4) \ dx \ \ ... (star)#

The first integral we can just evaluate directly,

# int \ 1/(x-2) \ dx = ln|x-2| #

We can deal with the second integral by splitting into two fractions:

# int \ (5x-6)/(x^2+4) \ dx = int \ (5x)/(x^2+4) - 6/(x^2+4) \ dx #
# " " = 5/2int \ (2x)/(x^2+4) \ dx - int \ 6/(x^2+4) \ dx #
# " " = 5/2ln|x^2+4| - 6/2arctan(x/2) #

Substituting these three integral results into #(star)# we get:

# I = 5/8 ln|x-2| - 1/8{5/2ln|x^2+4| - 3arctan(x/2)} +C #
# \ \ = 5/8 ln|x-2| - 5/16 ln|x^2+4| - 3/8arctan(x/2) + C#