Is it true that $\int_{0}^{1}(1+x^{2})^{-1/2} = \log (1 + \sqrt{2})$?

Question

Since $$D^{-1} (1 + x^{2})^{-1/2} = \sinh ^{-1} (x) + C,$$ is it true that $$\sinh ^{-1} x + C \big|_{0}^{1} = \log (1 + \sqrt{2})?$$ What relates $\sinh^{-1}(\cdot )$ to $\log(\cdot )$?

Here $D^{-1}$ is the primitive finder.

Answer 1

$x = \sinh t = \frac{1}{2}(e^t-e^{-t})$, then how to express $t$ by $x$? Solve the quadratic equation $$e^{2t} - 2xe^{t} - 1= 0 .$$

Answer 2

A yet different approach:

Let $x=\tan y$ to obtain \begin{align} \int_0^1\frac{dx}{\sqrt{1+x^2}}&=\int_0^{\frac{\pi}{4}}\sec y dy\\ &=\int_0^{\frac{\pi}{4}}\sec y \frac{\sec y + \tan y}{\sec y + \tan y}dy\\ &=\int_0^{\frac{\pi}{4}} \frac{\sec^2 y + \sec y \tan y}{\sec y + \tan y}dy\\ &=\int_0^{\frac{\pi}{4}} \frac{d(\sec y + \tan y)}{\sec y + \tan y}\\ &=\log|\sec y + \tan y|\Big|_{y=0}^{y=\frac{\pi}{4}}\\ &=\log|\sec \frac{\pi}{4} + \tan \frac{\pi}{4}|-\log|\sec 0 + \tan 0|\\ &=\log|\sqrt{2} + 1|-\log|1 + 0|=\log(\sqrt{2} + 1)\\ \end{align}