`int8/((x+2)(x^2+4))dx`

Explanation

Let ` I = int(8)/((x+2)(x^(2) +4))dx`
`(8)/((x+2)(x^(2)+4)) = (A)/(x+2) +(Bx +C)/(x^(2) + 4)`
` 8 = A(x^(2) + 4) +(Bx + C) (X+2) " ".......(1)`
`8 = (A +B)x^(2) +(2B +C) x + (4A +2C) " ".....(2)`
Putting x =- 2 in equation (1), we get
` 8 = 8 A rArr A = 1`
Comparing coefficiet of `x^(2)` and constant term on both sides, we get
`A + b = 0 and 4A + 2C = 8`
`B = - 1 and 4 + 2C = 8 rArr C = 2`
`therefore(8)/((x+2)(x^(2) +4)) = (1)/(x+2) +(-x +2)/(x^(2) + 4)`
`I = int[(1)/(x+2) +(-x +2)/(x^(2) + 4)] dx`
` = int(1)/(x+2) dx - (1)/(2)int(2x)/(x^(2) + 4) dx + 2 int(1)/(x^(2) +2^(2)) dx`
` = log|x +2| - (1)/(2) log |x^(2) + 4|+2.(1)/(2) tan^(-1) ((X)/(2)) + c`
`therefore int(8)/((x+2)(x^(2) + 4)) dx = log |(x+2)/(sqrt(x^(2) + 4))| + tan^(-1) ((x)/(2)) + c`