# Integral using trig substitution check

I am trying to integrate this:

$$\int_0^2 \frac{dt}{\sqrt{4 + t^2}}$$

after letting $$t = 2 \tan {\theta}$$, I find that the equation reduces to:

$$\int \sec {\theta}\, d \theta = \ln | \sec {\theta} + \tan {\theta} |$$ and I find that the indefinite integral after replacing the thetas is is:

$$\ln \Bigg| \frac{\sqrt{t^2 + 4}}{2} + t \,\Bigg|$$

when I take the definite integral I get:

$$\ln \big|\, 2 + \sqrt{2} \,\big|$$ which is not what wolfram gets:

What am I doing wrong?

$$\ln\left|\frac{1}{2}\sqrt{t^2+4} + \frac{t}{2}\right|$$
This gives $$\ln(1+\sqrt{2}) = \sinh^{-1}(1)$$.