Decompose: 8/(( x + 2 ) ( x^2 + 4 ))
8/(( x + 2 ) ( x^2 + 4 )) = A/(x+2)+ (Bx)/(x^2+4) + C/(x^2+4)
Multiply both sides by ( x + 2 ) ( x^2 + 4 ):
8 = A( x^2 + 4 ) + Bx( x + 2 ) + C(x + 2)
Let x = -2:
8 = A( (-2)^2 + 4 ) + B(-2)( -2 + 2 ) + C(-2 + 2)
8 = A(8) + B(-2)(0) + C(0)
A = 1
4 - x^2 = Bx( x + 2 ) + C(x + 2)
Let x = 0
4-0^2 = B(0)( 0 + 2 ) + C(0 + 2)
4 = 2C
C = 2
4 - x^2 = Bx( x + 2 ) + 2x + 4
- x^2 -2x= Bx( x + 2 )
B = -1
int8/(( x + 2 ) ( x^2 + 4 ))dx = int1/(x+2)dx- intx/(x^2+4)dx + int2/(x^2+4)dx
The first integral is the natural logarithm:
int8/(( x + 2 ) ( x^2 + 4 ))dx = ln|x+2|- intx/(x^2+4)dx + int2/(x^2+4)dx
Multiple the second integral by 2/2 and it, too, becomes a natural logarithm but without an absolute value, because the argument can never be negative:
int8/(( x + 2 ) ( x^2 + 4 ))dx = ln|x+2|- 1/2ln(x^2+4) + int2/(x^2+4)dx
The last integral is our old friend the inverse tangent:
int8/(( x + 2 ) ( x^2 + 4 ))dx = ln|x+2|- 1/2ln(x^2+4) + tan^-1(x/2)+C