Question

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

Answer 1

I=$\int \frac{x}{x^2+x+1}dx=\int \frac{\frac{1}{2}(2x+1)-\frac{1}{2}}{x^2+x+1}dx=\frac{1}{2}\int \frac{2x+1}{x^2+x+1}dx\left(i_1\right)-\frac{1}{2}\int \frac{1}{x^2+x+1}dx\left(i_2\right)+C$

$i_1$ can be read as $\frac{1}{2}\int \frac{dt}{t}$ for t=$x^2+x+1$

so $i_1=\frac{1}{2}ln|x^2+x+1|+c$

$i_2$ can simply be evaluated by making perfect square in denominator

$i_2=\frac{1}{2}\int \frac{1}{(x+\frac{1}{2}{)}^2+\frac{3}{4}}=\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+c$

So I=$i_1-i_2=$ $\frac{1}{2}ln|x^2+x+1|-\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C$