Question

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

Answer 1

`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`