I know the answer is $x - \ln|x^2 + x + 1|$ but I don't understand how to get it. Its in the partial fraction decomposition section of homework. The way the homework worked it is like this...
$$ \int \frac{2x+1}{x^2 + x + 1} \, dx $$
I see they split it up and got the derivative of $x^2 + x + 1$ in the numerator, but I don't understand how they did that or where the other interal dx came from.
Note that $(x^2 + x + 1) - (2x + 1) = x^2 - x$. So we can rewrite as
$$\int dx \left(1 - \frac{2x + 1}{x^2 + x + 1}\right)$$
Using $u = x^2 + x + 1$
$$x - \ln |x^2 + x + 1| + C$$
Suppose you have $$ \frac{\text{some polynomial}}{x^2 + x + 1}. $$ Do long division and get $$ \text{some polynomial} + \frac{\text{some first-degree polynomial}}{x^2+x+1}. $$ In this case you get $$ 1 + \frac{-2x-1}{x^2+x+1}. $$ First degree is the derivative of second degree, so $\dfrac{d}{dx}(x^2+x+1) = 2x+1$. So \begin{align} u & = x^2 + x + 1, \\[6pt] du & = (2x+1)\,dx. \end{align} Then we have $$ \int \left(1 - \frac 1 u \right) du. $$
Somewhat more typically, one might have something like $$ \frac{3x+5}{x^2+x+1}. $$ Then one would write $$ 3x+5 = \frac 3 2 \left(2x + \frac{10}3\right)= \frac 3 2 \left((2x+1) + \frac 7 3\right) $$ so we get $$ \int\frac{3x+5}{x^2+x+1} \,dx=\frac 3 2 \int \frac{du}{u} + \frac 7 2 \int \frac{dx}{x^2+x+1}. $$ That last integral would then need to get done by a different method, whose first step is completing the square.
