Decompose: 8/(( x + 2 ) ( x^2 + 4 ))8(x+2)(x2+4)
8/(( x + 2 ) ( x^2 + 4 )) = A/(x+2)+ (Bx)/(x^2+4) + C/(x^2+4)8(x+2)(x2+4)=Ax+2+Bxx2+4+Cx2+4
Multiply both sides by ( x + 2 ) ( x^2 + 4 )(x+2)(x2+4):
8 = A( x^2 + 4 ) + Bx( x + 2 ) + C(x + 2)8=A(x2+4)+Bx(x+2)+C(x+2)
Let x = -2x=−2:
8 = A( (-2)^2 + 4 ) + B(-2)( -2 + 2 ) + C(-2 + 2)8=A((−2)2+4)+B(−2)(−2+2)+C(−2+2)
8 = A(8) + B(-2)(0) + C(0)8=A(8)+B(−2)(0)+C(0)
A = 1A=1
4 - x^2 = Bx( x + 2 ) + C(x + 2)4−x2=Bx(x+2)+C(x+2)
Let x = 0x=0
4-0^2 = B(0)( 0 + 2 ) + C(0 + 2)4−02=B(0)(0+2)+C(0+2)
4 = 2C4=2C
C = 2C=2
4 - x^2 = Bx( x + 2 ) + 2x + 44−x2=Bx(x+2)+2x+4
- x^2 -2x= Bx( x + 2 )−x2−2x=Bx(x+2)
B = -1B=−1
int8/(( x + 2 ) ( x^2 + 4 ))dx = int1/(x+2)dx- intx/(x^2+4)dx + int2/(x^2+4)dx∫8(x+2)(x2+4)dx=∫1x+2dx−∫xx2+4dx+∫2x2+4dx
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∫8(x+2)(x2+4)dx=ln|x+2|−∫xx2+4dx+∫2x2+4dx
Multiple the second integral by 2/222 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∫8(x+2)(x2+4)dx=ln|x+2|−12ln(x2+4)+∫2x2+4dx
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∫8(x+2)(x2+4)dx=ln|x+2|−12ln(x2+4)+tan−1(x2)+C
