Question

Integrate the function: $\frac{1}{9x^2+6x+5}$

Answer 1

To find $\int \frac{1}{9x^2+6x+5}.dx$

$=\int \frac{1}{9(x^2+\frac{6x}{9}+\frac{5}{9})}.dx$

$=\frac{1}{9}\int \frac{1}{x^2+\frac{2x}{3}+{\left(\frac{1}{3}\right)}^2-{\left(\frac{1}{3}\right)}^2+\frac{5}{9}}.dx$

$=\frac{1}{9}\int \frac{1}{{(x+\frac{1}{3})}^2-\frac{1}{9}+\frac{5}{9}}.dx$

$=\frac{1}{9}\int \frac{1}{{(x+\frac{1}{3})}^2+\frac{4}{9}}.dx$

$=\frac{1}{9}\int \frac{1}{{(x+\frac{1}{3})}^2+{\left(\frac{2}{3}\right)}^2}.dx$

$=\frac{1}{9}\left[\frac{1}{\frac{2}{3}}{\tan }^{-1}\left(\frac{(x+\frac{1}{3})}{\frac{2}{3}}\right)\right]+C$

$[\because \int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C]$

$=\frac{1}{9}\left[\frac{3}{2}{\tan }^{-1}\left(\frac{3x+1}{2}\right)\right]+C$

$=\frac{1}{6}{\tan }^{-1}\left(\frac{3x+1}{2}\right)+C$

where $C$ is a constant of integration .