Question

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

Answer 1

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

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

$=\frac{1}{2}\int \frac{1}{x^2+\frac{x}{2}+\frac{1}{16}-\frac{1}{16}+\frac{1}{2}}dx$

$=\frac{1}{2}\int \frac{1}{{\left(x+\frac{1}{4}\right)}^2-\frac{1}{16}+\frac{1}{2}}dx$

$=\frac{1}{2}\int \frac{1}{{\left(x+\frac{1}{4}\right)}^2+\frac{7}{16}}dx$

We know that, $\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C$

$=\frac{1}{2}\left(\frac{1}{\frac{\sqrt{7}}{4}}{tan}^{-1}\left[\frac{x+\frac{1}{4}}{\frac{\sqrt{7}}{4}}\right]\right)+c$

$I=\frac{2}{\sqrt{7}}{tan}^{-1}\left[\frac{(4x+1)}{\sqrt{7}}\right]+c$

Hence,

$\int \frac{1}{{2x}^2+x+1}dx$$=\frac{2}{\sqrt{7}}{tan}^{-1}\left[\frac{(4x+1)}{\sqrt{7}}\right]+c$