Question #15823

Sep 7, 2017

What can we try? We can try factoring out. If $N = D \cdot P$ then $\frac{N}{D} = P$ where $P$ is a polynomial. We know how to integrate polynomials.

Explanation:

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

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