Question

Evaluate the integral `int \ (4x^3-7x)/(x^4-5x^2+4) \ dx`?

Answer 1

`int \ (4x^3-7x)/(x^4-5x^2+4) \ dx = 3/2 ln|x^2-4| + 1/2 ln|x^2-1| + c`

Explanation:

Denote the integral by:

`I = int \ (4x^3-7x)/(x^4-5x^2+4) \ dx`

The denominator of the integrand is quadratic in `x^2` and factorises as follows:

`(4x^3-7x)/(x^4-5x^2+4) -= (4x^3-7x)/((x^2-4)(x^2-1))`
`" " = (4x^3-7x)/((x^2-2^2)(x^2-1^2))`
`" " = (4x^3-7x)/((x-2)(x+2)(x-1)(x+1))`

And so the partial fraction decomposition will be of the form:

`(4x^3-7x)/(x^4-5x^2+4) -= A/(x-2) + B/(x+2) + C/(x-1) + D/(x-1)`

Then if we put the RHS over a common denominator we end up with the identity:

`(4x^3-7x) -= A(x+2)(x-1)(x+1) + B(x-2)(x-1)(x+1) + C(x-2)(x+2)(x+1) + D(x-2)(x+2)(x-1)`

Where `A,B,C,D` are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put `x= \ \ \ \ 2 => 4*8-14 = A(4)(1)(3) => A=3/2`
Put `x=-2 => -4*8+14 = B(-4)(-3)(-1) => B=3/2`
Put `x= \ \ \ \ 1 => 4-7 = C(-1)(3)(2) => C=1/2`
Put `x=-1 => -4+7 = D(-3)(1)(-2) => D = 1/2`

So using partial fraction decomposition we have:

`I = int \ (3/2)/(x-2) + (3/2)/(x+2) + (1/2)/(x-1) + (1/2)/(x-1) \ dx`

These are all trivial integrals, so we can integrate to get:

`I = 3/2 ln|x-2| + 3/2 ln|x+2| + 1/2 ln|x-1| + 1/2 ln|x-1| + c`
`\ \ = 3/2 (ln|x-2| + ln|x+2|) + 1/2 (ln|x-1| +ln|x-1|) + c`
`\ \ = 3/2 (ln|x-2||x+2|) + 1/2 (ln|x-1||x-1|) + c`
`\ \ = 3/2 (ln|(x-2)(x+2)|) + 1/2 (ln|(x-1)(x-1)|) + c`
`\ \ = 3/2 ln|x^2-4| + 1/2 ln|x^2-1| + c \ \ \ \ \` QED