On the integral $\int_0^1\frac{1}{1+x^2}dx$

Question

It is easy to show through the substitution $x = \sin\varphi$ that the integral

$$\int_0^1 \frac{1}{1+x^2}\ dx = \frac \pi 4.$$

I have just attempted this integral employing the hyperbolic substitution $x = \sinh\varphi,$ prompting the integral $$\int_{x = 0}^{x = 1}\operatorname{sech}\varphi \ d\varphi,$$ which after evaluating and inputting $x$, I find

$$\int_{x = 0}^{x = 1}\operatorname{sech}\varphi \ d\varphi = \ln \left| \frac{3\pm\sqrt{2} - \frac 1 {1\pm\sqrt{2}}}{1\pm\sqrt{2} + \frac{1}{1\pm\sqrt{2}}} \cdot\frac{1\pm1}{2}\right|.$$

To avoid an infinite answer, I take the positive solution, yielding

$$\int_{x=0}^{x=1}\operatorname{sech}\varphi\ d\varphi = \ln \left( \sqrt 2 \right).$$

Clearly

$$\frac \pi 4 \neq \ln\left(\sqrt 2 \right).$$

I'm confident in my algebra in obtaining this result (though I can certainly include all my steps if needed). Hence, I am curious as to why the hyperbolic substitution does not yield the proper answer.

Answer 1

$$\int_{x=0}^{x=1} \operatorname{sech}(u) \, du = \left[ 2\tan^{-1} \left( \tanh \left( \frac u 2 \right)\right)\right]_{\operatorname{arsinh}(0)}^{\operatorname{arsinh}(1)} = \frac \pi 4 \text{ as required} $$

Answer 2

$x = \sinh \varphi$ means $\varphi = \log\left(x+\sqrt{1+x^2}\right)$. Therefore when $x$ varies between $0$ and $1$, $\varphi$ varies between $0$ and $\log(1+\sqrt 2)$.

Therefore,

$$\int_0^1 \frac 1 {1+x^2} \, dx = \int_0^{\log(1+\sqrt 2)}\operatorname{sech} \varphi \, d \varphi = \left[ 2 \arctan \left(e^\varphi\right) \vphantom{\frac11} \right]_0^{\log \left(1+\sqrt 2\right)} = \frac \pi 4 $$