Integration Partial Fractions

Question

I have the following problem: $$\int\left(\frac{x+2}{x^2+x+1}\right)dx$$ I received this by simplification of another integral. But my question is how to procede from here. Is there a way to simplify this?

Answer 1

Hint:

You can re-write the integral as \begin{align*} \int\frac{x+\frac{1}{2}}{x^2+x+1}dx+\int\frac{\frac{3}{2}}{x^2+x+1}dx&=\frac{1}{2}\int\frac{d(x^2+x+1)}{x^2+x+1}+\frac{3}{2}\int\frac{1}{(x+\frac12)^2+\frac34}dx \end{align*}

Answer 2

Partial fractions doesnt work here since $x^2+x+1$ is irreducible. Try separating into

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

The left is a simple $u$ substitution and the right looks like arctan if you complete the square in the denominator.

Answer 3

$$\int\frac{x+2}{x^2+x+1}dx=\frac12\int\frac{2x+1}{x^2+x+1}dx+\frac32\int\frac{dx}{\frac34+\left(x+\frac12\right)^2}=$$

$$=\frac12\log(x^2+x+1)+2\int\frac{dx}{1+\left(\frac{2x+1}{\sqrt3}\right)^2}=$$

$$=\log\sqrt{x^2+x+1}+\sqrt3\,\arctan\frac{2x+1}{\sqrt3}+K$$

Answer 4

The denominator doesn't factor. In fact, the roots are complex. Partial fractions aren't going to get you anywhere. Complete the square, and do a tan substitution.