Question

Evaluate $\int \frac{dx}{9x^2+6x+5}$

Answer 1

$\int \frac{dx}{9x^2+6x+5}$

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

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

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

$=\frac{1}{9}\times \frac{3}{2}{\tan }^{-1}\cdot \frac{\left(x=\frac{1}{3}\right)}{\left(\frac{2}{3}\right)}+c$

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