$\int \frac{\sqrt{4x^2-8x+3}}{x-1}dx$ Integrate $$\int \frac{\sqrt{4x^2-8x+3}}{x-1}dx$$
What I first did is I tried to complete the square for the numerator
$\sqrt{4(x^2-2x+3/4)} = \sqrt{4(x^2-2x+1-1/4)} = \sqrt{4(x-1)^2 - 1}$
Now I did trig-substitution:
$(x-1) = \frac{1}{2}\sec(\theta)$
$dx = \frac{1}{2}\sec(\theta)\tan(\theta)d\theta$
$$=\int \frac{\tan^2(\theta)\sec(\theta)}{\frac{1}{2}\sec(\theta) }d\theta = 2\tan(\theta) - 2\theta + C = 2\frac{\sqrt{4x^2-8x+3}}{2} - 2\sec^{-1}\left(\frac{x-1}{2}\right) + C$$

I think your last step is wrong.
$2\tan\theta - 2\theta + c$
$= 2 \sqrt {\sec^2 \theta - 1} - 2 \theta + c$
As we have,
$\sec \theta = 2(x - 1)$
$\theta = \sec^{-1}2(x - 1)$
Put these values above,
$= 2 \sqrt {4x^2-8x+3} - 2 \sec^{-1}2(x - 1)+ c$
