Fractional Integration by substitution

Question

The first part obviously indicates to simplify the question down to make it easier to look at

Thus by using (x-1) as a factor of x^3 -1 I arrive as formula (2x+1)/(x^2 +1)

To integrate that you separate it out to two fractions and easily integrate each term to arrive at

$$2ln|x^2+1| + arctan(x) + c$$ but the book says the answer is $$ln|x^2+x+1| + c$$ Which doesn't make sense

Answer 1

Your mistake was you forgot the $x$ term in $x^2+x+1$ when you factored $x^3-1$.

We have this:

$$\int\frac{2x^2-x-1}{x^3-1}dx$$

Factoring out $x-1$:

$$=\int\frac{2x+1}{x^2+x+1}dx$$

Letting $u=x^2+x+1$ and $du=2x+1\;dx$:

$$=\int\frac{1}{u}du$$

Integrating and plugging $u$ back in:

$$=\ln|x^2+x+1|+C$$

Answer 2

Hint:

$x^3 - 1 = (x-1)(x^2+x+1)$

Note:

$$2x^2 -x-1 = (x-1)(2x+1)$$

So your integrand becomes:

$$ \frac{2x+1}{x^2+x+1}$$

Integrating the above is straight forward:

$$ \frac{d}{dx}(x^2+x+1) = 2x + 1$$