Question

How do you integrate `f(x)=(-x^2-2x)/((x^2+2)(x+7))` using partial fractions?

Answer 1

`-8/51ln(x^2+2) + (5sqrt2)/51arctan(x/sqrt2)-35/51ln abs(x+7)+C`

Explanation:

`(-x^2-2x)/((x^2+2)(x+7)) = (Ax+B)/(x^2+2)+C/(x+7)=`

`=((Ax+B)(x+7))/((x^2+2)(x+7))+(C(x^2+2))/((x^2+2)(x+7))=`

`= (Ax^2+7Ax+Bx+7B+Cx^2+2C)/((x^2+2)(x+7))=`

`= ((A+C)x^2 +(7A+B)x + 7B+2C)/((x^2+2)(x+7))`

`(-x^2-2x)/((x^2+2)(x+7)) = ((A+C)x^2 +(7A+B)x + 7B+2C)/((x^2+2)(x+7))`

`A+C = -1 => 7A+7C = -7`
`7A+B = -2`
`7B+2C=0`

`B-7C = 5 = > 7B-49C=35`
`7B+2C=0`

`51C=-35 => C = -35/51`

`B=5+7C = 5 - 245/51 = 10/51`

`A = -1-C = -1 +35/51 = -16/51`

`I = int (-x^2-2x)/((x^2+2)(x+7))dx = int (-16/51x+10/51)/(x^2+2)dx + int (-35/51)/(x+7)dx`

`I = -16/51int(xdx)/(x^2+2)+10/51intdx/(x^2+2)-35/51intdx/(x+7)`

`I = -16/51int (1/2d(x^2+2))/(x^2+2)+10/51int dx/(x^2+(sqrt2)^2)-35/51intdx/(x+7)`

`I = -8/51ln(x^2+2) + (5sqrt2)/51arctan(x/sqrt2)-35/51ln abs(x+7)+C`