Question

Integrate

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

Answer 1

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

$\Rightarrow \int \frac{\frac{1}{2}(2x+1)-1/2}{x^2+x+1}dx$

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

Putting $x^2+x+1=t$

$(2x+1)dx=t$

$\Rightarrow \frac{1}{2}\int \frac{dt}{t}=\frac{1}{2}\int \frac{dx}{x^2+x+\frac{1}{4}+\frac{3}{4}}$

$=\frac{1}{2}ln\left|t\right|-\frac{1}{2}\int \frac{dx}{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$

$\Rightarrow \frac{1}{2}ln\left|t\right|-\frac{1}{2}{tan}^{-1}\left(\frac{x+1/2}{\sqrt{3}/2}\right)\times \frac{2}{\sqrt{3}}+C$

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