The integral $\int \frac{2x-3}{{\left(x^2+x+1\right)}^2}$ dx is equal to 

Solution:

We have, 

$\begin{array}{l} I=\int \frac{2x-3}{{\left(x^2+x+1\right)}^2}dx=\int \frac{(2x+1)-4}{{\left(x^2+x+1\right)}^2}dx\\ \Rightarrow I=\int \frac{2x+1}{{\left(x^2+x+1\right)}^2}dx-4\int \frac{1}{{\left[(x+1/2{)}^2+(\sqrt{3}/2{)}^2\right]}^2}dx\\ I=-\frac{1}{x^2+x+1}-4I_1\\ \text{ where } I_1=\int \frac{1}{{\left[\right(x+1/2{)}^2+(\sqrt{3}/2{)}^2]}^2}dx\\ \end{array}$

Putting $x+1/2=(\sqrt{3}/2) \tan \theta \text{ in }{I}_1\text{, }$We get

$\begin{array}{l} I_1=\int \frac{\sqrt{3}/2{\sec }^2\theta d\theta }{{\left[\left[\right(\sqrt{3}/2)\tan \theta ){]}^2+(\sqrt{3}/2{)}^2\right]}^2}\\ \Rightarrow I_1=\frac{8}{3\sqrt{3}}\int {\cos }^2\theta d\theta =\frac{8}{3\sqrt{3}}\int \frac{1+\cos 2\theta }{2}d\theta \\ \Rightarrow I_1=\frac{4}{3\sqrt{3}}\{\theta +\frac{1}{2}\sin 2\theta \}+C\\ \Rightarrow I_1=\frac{4}{3\sqrt{3}}\left\{{\tan }^{-1}\frac{2x+1}{\sqrt{3}}+\frac{1}{2}\frac{2\tan \theta }{1+{\tan }^2\theta }\right\}+C\\ \Rightarrow I_1=\frac{4}{3\sqrt{3}}\left\{{\tan }^{-1}\frac{2x+1}{\sqrt{3}}+\frac{\sqrt{3}}{4}\left(\frac{2x+1}{x^2+x+1}\right)\right\}+C\\ \therefore I=-\frac{1}{x^2+x+1}-\frac{16}{3\sqrt{3}}{\tan }^{-1}\frac{2x+1}{\sqrt{3}}-\frac{4}{3}\left(\frac{2x+1}{x^2+x+1}\right)+C\end{array}$