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#
