$\int \frac{1}{\sqrt{3-6x+9x^2}} \, dx$ is equal to
$\sin^{-1}\left(\frac{3x+1}{2}\right) + c$
$\sin^{-1}\left(\frac{3x+1}{6}\right) + c$
$\frac{1}{3} \sin^{-1}\left(\frac{3x+1}{2}\right) + c$
$\sin^{-1}\left(\frac{2x+1}{3}\right) + c$
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