Question

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

Answer 1

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

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

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

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

take $3x+1=t$

$\Rightarrow 3dx=dt$

Substitute this in $\left(1\right)$, we get

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

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

This is in the form of $\int \frac{1}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C$

$\Rightarrow \frac{1}{3}\times \frac{1}{2}{\tan }^{-1}\left(\frac{t}{2}\right)+C$

but $t=3x+1$

Substituting the value of $t$ in the above expression, we get

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

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