Question

Question #15823

Answer 1

What can we try? We can try factoring out. If `N=D*P` then `N/D=P` where `P` is a polynomial. We know how to integrate polynomials.

Explanation:

Let's try factoring out. We can call the denominator D
`D = x^6 - x^4 = (x^4)(x^2-1) = x^4(x-1)(x+1)`
The numerator N is `x^5 + 1`. It is zero for `x=-1` so it contains the factor `(x-(-1)) = x+1`
Namely `N=x^5 + 1 = (x+1)(x^4-x^3+x^2-x+1)`
Now we are ready to divide
`N/D = (x^4-x^3+x^2-x+1)1/(x^4)(x-1)`

Now if the last term `1` wasn't there on the numerator, we could divide in a straightforward way. So we rewrite `N` as
`N = N -1 + 1`
`N = x^3(x-1) + x(x-1) + 1`
and we obtain
`N/D = \frac{1}{x} + \frac{1}{x^3} + \frac{1}{x^4(x-1)}`
`N/D=\frac{1}{x} + \frac{1}{x^3} - \frac{x^3+x^2+x+1}{x^4}+\frac{1}{x-1}`
and in the end
`N/D = -\frac{1}{x^2} - \frac{1}{x^4} +\frac{1}{x-1}`
And
`I = \frac{1}{x} -\frac{1}{2x^2}+ log|x-1|`