Question

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

Answer 1

$\frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + C$

Answer 2

To evaluate the integral $\int \frac{dx}{x^2+x+1}$, we need to complete the square in the denominator.

The denominator is $x^2+x+1$. To complete the square for a quadratic expression of the form $ax^2+bx+c$, we can write it as $a\left(x+\frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}$. In this case, $a=1$, $b=1$, $c=1$. So, $x^2+x+1 = \left(x+\frac{1}{2}\right)^2 + 1 - \frac{1^2}{4(1)}$ $= \left(x+\frac{1}{2}\right)^2 + 1 - \frac{1}{4}$ $= \left(x+\frac{1}{2}\right)^2 + \frac{4-1}{4}$ $= \left(x+\frac{1}{2}\right)^2 + \frac{3}{4}$

Now, substitute this back into the integral: $\int \frac{dx}{x^2+x+1} = \int \frac{dx}{\left(x+\frac{1}{2}\right)^2 + \frac{3}{4}}$ This integral is of the form $\int \frac{du}{u^2+a^2}$, where $u = x+\frac{1}{2}$ and $a^2 = \frac{3}{4}$. If $u = x+\frac{1}{2}$, then $du = dx$. And $a = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

Using the standard integral formula $\int \frac{du}{u^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$: $\int \frac{dx}{\left(x+\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + C$ Simplify the expression: $= \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{\frac{2x+1}{2}}{\frac{\sqrt{3}}{2}}\right) + C$ $= \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + C$

Explanation of the solution:

1. Complete the square: Rewrite the quadratic denominator $x^2+x+1$ as $(x+1/2)^2 + 3/4$.
2. Identify standard form: The integral transforms to $\int \frac{dx}{(x+1/2)^2 + (\sqrt{3}/2)^2}$, which is in the form $\int \frac{du}{u^2+a^2}$.
3. Apply formula: Use the standard integral formula $\int \frac{du}{u^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$ with $u = x+1/2$ and $a = \sqrt{3}/2$.
4. Simplify: Substitute and simplify the expression to get the final result.