Using Inverse Trig. Technique $\int$$\frac{dx}{\sqrt{(x-1)^2+4}}$
Let $x = 2\tan\theta +1$
$dx= 2\sec^2\theta d\theta$
$= \int$$\frac{2\sec^2\theta d\theta}{\sqrt{4\sec^2\theta}}$
$=\int\sec d\theta$
$= \ln(\sec\theta + \tan\theta) + C$
Triangle based off of x substitution: opposite= $x-1$ 
adjacent=2 and hypotenuse = $\sqrt{(x-1)^2+4}$
Thus answer $=\ln(\frac{\sqrt{(x-1)^2+4}}{2} + \frac{x-1}{2}) + C$
However according to my given solution it should be equal to
$=\ln(\sqrt{x^2-2x+5} + x-1) + C$
I know that this is unfactored form of the polynomial but i do not know why the 
2 in my denominator is incorrect.

$$\ln\left(\frac{\sqrt{(x-1)^2+4}}{2} + \frac{x-1}{2}\right) + C=\ln\left(\frac{\sqrt{(x-1)^2+4}+x-1}{2}\right) + C=\ln\left(\sqrt{(x-1)^2+4}+x-1\right)-\ln 2 + C=\ln\left(\sqrt{(x-1)^2+4}+x-1\right)+C'$$
You should know $\ln ab=\ln a+\ln b$.
