Question

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

Answer 1

We are write denominator as

$1+x+x^2=1+\frac{1}{4}+2.\frac{1}{2}.x+x^2-\frac{1}{4}=\frac{3}{4}+{(\frac{1}{2}+x)}^2$

$\therefore \int \frac{1}{1+x+x^2}dx=\int \frac{1}{\frac{3}{4}+{(x+\frac{1}{2})}^2}dx$

$=\int \frac{1}{\frac{3}{4}[1+\frac{4}{3}{(x+\frac{1}{2})}^2]}dx=\frac{4}{3}\int \frac{1}{1+{[\frac{2}{\sqrt{3}}+(x+\frac{1}{2})]}^2}dx$

$=\frac{4}{3}\int \frac{1}{1+{(\frac{2}{\sqrt{3}}x+\frac{1}{\sqrt{3}})}^2}dx=\frac{4}{3}.\frac{\sqrt{3}}{2}\int \frac{\frac{2}{\sqrt{3}}}{1+{(\frac{2x}{\sqrt{3}}x+\frac{1}{\sqrt{3}})}^2}dx$

$=\frac{2}{\sqrt{3}}arc\tan (\frac{2x}{\sqrt{3}}x+\frac{1}{\sqrt{3}})=\frac{2}{\sqrt{3}}arc\tan \left(\frac{2x+1}{\sqrt{3}}\right)$Ans.