Question

How do you use partial fraction decomposition to decompose the fraction to integrate `1/((x+2)(x^2+4))`?

Answer 1

`I = 1/8ln|x+2| - 1/16 ln(x^2+4) + 1/8arctan(x/2) +C`

Explanation:

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

`= (A(x^2+4))/((x+2)(x^2+4))+((Bx+C)(x+2))/((x^2+4)(x+2)) =`

`= (Ax^2+4A+Bx^2+2Bx+Cx+2C)/((x+2)(x^2+4)) =`

`= ((A+B)x^2+(2B+C)x+4A+2C)/((x+2)(x^2+4))`

`(A+B)x^2+(2B+C)x+4A+2C = 1`

`A+B=0 => A=-B`
`2B+C=0 => C=-2B`
`4A+2C=1 => 4(-B)+2(-2B)=1`

`4(-B)+2(-2B)=1`

`-8B=1 => B=-1/8`

`A=-B => A=1/8`

`C=-2B => C=1/4`

`I = int 1/((x+2)(x^2+4)) dx = 1/8int 1/(x+2)dx + int (-1/8x+1/4)/(x^2+4)dx`

`I = 1/8ln|x+2| - 1/8 int (xdx)/(x^2+4) + 1/4int dx/(x^2+4)`

`I = 1/8ln|x+2| - 1/16 int ((x^2+4)'dx)/(x^2+4) + 1/4 *1/2arctan(x/2)`

`I = 1/8ln|x+2| - 1/16 int (d(x^2+4))/(x^2+4) + 1/8arctan(x/2)`

`I = 1/8ln|x+2| - 1/16 ln(x^2+4) + 1/8arctan(x/2) +C`