Question

Solve:

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

Answer 1

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

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

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

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

$=\frac{1}{2}log(x^2+x+1)-\frac{1}{2}\int \frac{1}{x^2+2.x\frac{1}{2}+(\frac{1}{2}{)}^2-(\frac{1}{2}{)}^2+1}$

$=\frac{1}{2}log(x^2+x+1)-\frac{1}{2}\int \frac{1}{(x+\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2}dx$

Let $x+\frac{1}{2}=t\Rightarrow dx=dt$

$=\frac{1}{2}log(x^2+x+1)-\frac{1}{2}\int \frac{dt}{t^2+(\frac{\sqrt{3}}{2}{)}^2}$

$=\frac{1}{2}log(x^2+x+1)-\frac{1}{2}.\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}te{n}^{-1}\left(\frac{t}{\left(\sqrt{3}/2\right)}\right)+c$

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

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