Question

Find $\int \frac{1}{9x^2+6x+5}dx$

Answer 1

$\int \frac{dx}{a{x}^2+6x+5}=\frac{1}{9}\int \frac{dx}{x^2+\frac{2x}{3}+\frac{5}{9}}$

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

$=\frac{1}{9}\int \frac{dx}{(x+\frac{1}{3}{)}^2+(\frac{2}{3}{)}^2}$ $[\because \int \frac{dx}{x^2+a^2}$ = $\frac{1}{a}ta{n}^{-1}\left(\frac{x}{a}\right)+c]$

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

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