Question

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

Answer 1

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`