Question

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

Answer 1

${\displaystyle \frac{1}{9x^2+6x+5}}$

I = ${\displaystyle \int \frac{1}{9x^2+6x+5}dx=\int \frac{1}{{(3x+1)}^2+2^2}}$

let (3x + 1) = t ⇒ 3dx = dt

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

= ${\displaystyle \frac{1}{3}\int \frac{1}{t^2+2^2}dt}$

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

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

Substituting the value of t

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

Answer 2

Let ${\displaystyle I=\int \frac{dx}{9x^2+6x+5}}$

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

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

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

${\displaystyle =\frac{1}{9}\times \frac{1}{\frac{2}{3}}{\tan }^{-1}\left(\frac{x+\frac{1}{3}}{\frac{2}{3}}\right)+C}$             ${\displaystyle ....[\because \int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1} \frac{x}{a}+C]}$

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