Question

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

Answer 1

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

$Let \ (3x+1) = t$

$∴ 3dx = dt$

⇒ $∫\frac {1}{(3x+1)^2+(2)^2}dx$ $=$ $\frac 13 ∫\frac {1}{t^2+2^2} dt$

                                       $=\frac 13 [\frac 12\ tan^{-1}(\frac t2)]+C$

                                       $=\frac 16\ tan^{-1}(\frac {3x+1}{2})+C$