`(8)/((x+2)(x^(2)+4).)`

Explanation

`"Let I"=int (8)/((x+2)(x^(2)+4))dx`
`"Let "=(8)/((x+2)(x^(2)+4))=(A)/(x+2)+(Bx+C)/(x^(2)+4)`
`therefore 8=A(x^(2)+4)+(Bx+C)(x+2)`
`"Put x+2=0, i.e,x=-2, we get"`
`=8A(4+4)+(-2B+C)(0)`
`therefore 8=8A" "therefore A=1`
Put x=0, we get
8=A(4)+(0+C)(2)
`therefore 8=4A+2C" "therefore 4+2C therefore C=2`
Put x=1, we get
`=8=A(1+4)+(B+C)(1+2)`
`therefore 8=5A+3B+3C`
But A=1 and C=2
`therefore 8=5+3B+6" "therefore 3B=-3 " "therefore B=-1`
`therefore (8)/((x+2)(x^(2)+4))=(1)/(x+2)+((-x+2))/(x^(2)+4)`
`therefore I=int [(1)/(x+2)+(-x+2)/(4)]dx`
`=int (1)/(x+2)dx-(1)/(2)int (2x)/(x^(2)+4)dx+2int(1)/(x^2+2^2)dx`
`=log|x+2|-(1)/(2)log|x^(2)+4|+2xx(1)/(2)tan^(-1)((x)/(2))+c ...[because (d)/(dx)(x^(2)+4)=2x and int (f'(x))/(f(x))dx=log|f(x)|+c]`
`=log|x+2|-(1)/(2)log|x^(2)+4|+tan^(-1)((x)/(2))+c`