You should write the given fraction such that:

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

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

You need to factor out `x^3` in `x^5 + x^3` such that:

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

Reducing by `x^2 + 1` yields:

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

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

Substituting `(1/3)(x^3/(x^2 - 2) - x^3/(x^2+1))` for `x^3/((x^2+1)(x^2-2))` yields:

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

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

You may write the fraction `(x^3/(x^2 - 2))` such that:

`(x^3 - 2x + 2x)/(x^2 - 2) = (x^3 - 2x)/(x^2 - 2) + (2x)/(x^2 - 2) `

`(x^3 - 2x + 2x)/(x^2 - 2) = x(x^2 - 2)/(x^2 - 2) + (2x)/(x^2 - 2)`

`(x^3 - 2x + 2x)/(x^2 - 2) = x + (2x)/(x^2 - 2)`

You may write the fraction `(x^3)/(x^2+1)` such that:

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

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

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

You may evaluate the indefinite integral such that:

`int x^5/((x^2+1)(x^2-2)) dx= (2/3)int (x + (2x)/(x^2 - 2))dx + (1/3) int (x - x/(x^2+1)) dx`

`int x^5/((x^2+1)(x^2-2)) dx = int x dx + (2/3)int (2x)/(x^2 - 2)dx- (1/3) int x/(x^2+1) dx`

You should use the substitution to solve integrals `(2/3)int (2x)/(x^2 - 2)dx` and `(1/3) int x/(x^2+1) dx` such that:

`(2/3)int (2x)/(x^2 - 2)dx `

You should come up with th substitution such that:

`x^2 - 2 = y =gt 2xdx = dy`

Changing the variable yields:

`(2/3)int (dy)/y = (2/3) ln |y| + c `

`(2/3)int (2x)/(x^2 - 2)dx = (2/3) ln |x^2 - 2| + c`

`(1/3) int x/(x^2+1) dx `

You should come up with th substitution such that:

`x^2 + 1 = y =gt 2xdx = dy =gt xdx = (dy)/2 `

`(1/3) int x/(x^2+1) dx = (1/6) int (dy)/y`

`(1/6) int (dy)/y = (1/6) ln |y| + c`

`(1/3) int x/(x^2+1) dx =(1/6) ln (x^2+1) + c`

`int x^5/((x^2+1)(x^2-2)) dx = x^2/2 + (2/3) ln |x^2 - 2| - (1/6) ln (x^2+1) + c`

`int x^5/((x^2+1)(x^2-2)) dx = x^2/2 + ln (root(3)((x^2 - 2)^2)/(root(6)(x^2+1))) + c`

**Hence, evaluating the given integral yields `int x^5/((x^2+1)(x^2-2)) dx = x^2/2 + ln (root(3)((x^2 - 2)^2)/(root(6)(x^2+1))) + c.` **