Question

Solve $\frac{1}{3}\int \frac{x-1}{x^2+x+1}dx$

Answer 1

$=\frac{1}{3}\int \frac{x-1}{x^2+x+1}dx$

$\frac{1}{3}\left[\int \frac{x+\frac{1}{2}}{(x+\frac{1}{2}{)}^2+{\left(\sqrt{3}/2\right)}^2}dx\frac{-3}{2}\int \frac{dx}{(x+1/2{)}^2+(\sqrt{3}/2^2)}\right]$

Put $x+1/2=t$ $dx=dt$

$=\frac{1}{3}\left[\int \frac{tdt}{t^2+(\sqrt{3}/2{)}^2}-\frac{3}{2}\int \frac{dt}{t^2+(\sqrt{3}/2{)}^2}\right]$

Let put $t^2=u$ $2tdt=du$

$=\frac{1}{3}\left[\int \frac{du}{4+3/4}-\frac{3}{2}\times \frac{2}{\sqrt{3}}{\tan }^{-1}\frac{t\times 2}{\sqrt{3}}\right]$

$=\frac{1}{3}\left[\frac{1}{2}log|\right(x+1/2{)}^2+3/4|-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)]+c$

$=\frac{1}{6}log|x^2+x+1|-\frac{1}{3\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+c$