Question

How do you integrate `int(x(x+2))/(x^3 +3x^2 -4) dx`?

Answer 1

`1/3 ln|x-1|+ 2/3 ln|x+2|+C`

Explanation:

Let us begin by factorizing the denominator. It is easy to see that `x^3+3x^2-4` vanishes when `x=1`, so that `(x-1)` is a factor.

`x^3+3x^2-4 = x^3-x^2+4x^2-4x+4x-4`
`= x^2(x-1)+4x(x-1)+4(x-1) = (x-1)(x^2+4x+4) = (x-1)(x+2)^2`

Thus, the integrand simplifies considerably to

`{x(x+2)}/{x^3+3x^2-4} = {x(x+2)}/{(x-1)(x+2)^2} = x/{(x-1)(x+2)}`

We try the partial fraction expression

`x/{(x-1)(x+2)} = A/{x-1}+B/{x+2}`

Thus
`x = A(x+2) +B(x-1)`

substituting `x=1` and `x=-2`, respectively, gives

`1 = A times 3 implies A =1/3, -2 = B times (-3) implies B=2/3`

Thus

`x/{(x-1)(x+2)} = 1/3 1/{x-1}+2/3 1/{x+2}`

Finally

`int {x(x+2)}/{x^3+3x^2-4} dx = int ( 1/3 1/{x-1}+2/3 1/{x+2}) dx`
`qquad = 1/3 ln|x-1|+ 2/3 ln|x+2|+C`

Answer 2

`\qquad \qquad int \ { x ( x + 2 )}/{ x^3 + 3 x^2 - 4 } \ dx \ = \ 1/3 ln | x^3 + 3 x^2 - 4 | + C.`

Explanation:

`"We want to find:"`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad int \ { x ( x + 2 )}/{ x^3 + 3 x^2 - 4 } \ dx.`

`"Taking a quick second look, multiplying out the numerator,"`
`"we see:"`

`\qquad \qquad \qquad int \ { x ( x + 2 )}/{ x^3 + 3 x^2 - 4 } \ dx \ = \ int \ { x^2 + 2 x}/{ x^3 + 3 x^2 - 4 } \ dx.`

`"Observing the derivative of the denominator:" \qquad 3 x^2 + 6 x,`
`"which just happens to be proportional to numerator:" \ \ x^2 + 2 x,`
`"[this situation is a rare accident -- but welcome], we can finish"`
`"directly as:"`

`\qquad \qquad \qquad int \ { x ( x + 2 )}/{ x^3 + 3 x^2 - 4 } \ dx \ = \ int \ { x^2 + 2 x }/{ x^3 + 3 x^2 - 4 } \ dx.`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ 1/3 int \ { 3 ( x^2 + 2 x ) }/{ x^3 + 3 x^2 - 4 } \ dx.`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ 1/3 int \ { 3 x^2 + 6 x }/{ x^3 + 3 x^2 - 4 } \ dx.`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ 1/3 int \ { ( x^3 + 3 x^2 - 4 )' }/{ x^3 + 3 x^2 - 4 } \ dx.`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ 1/3 ln | x^3 + 3 x^2 - 4 | + C.`

`"Thus:"`

`\qquad \qquad \qquad int \ { x ( x + 2 )}/{ x^3 + 3 x^2 - 4 } \ dx \ = \ 1/3 ln | x^3 + 3 x^2 - 4 | + C.`