How do you integrate 8 / [ ( x + 2 ) ( x^2 + 4 ) ] using partial fractions?

1 Answer
Apr 15, 2017

Please see the explanation.

Explanation:

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