On $\int \frac{dx}{1+x^2}$

Question

I am well aware of the identity $$\tag{1} \int \frac{dx}{1+x^2}=\arctan x +C $$ and the derivation thereof (through trigonometric substitution). I attempted the derivation of $(1)$ through hyperbolic substitutions and found that of the two obvious substitutions---namely, $x=\sinh\phi$ or $x=\text{csch}\,\phi$---we arrive at an integral that is solved via the exact same trigonometric technique initially used to derive $(1)$. For instance, the first substitution yields $$\int \text{sech}\, \phi\, d\phi \implies \int \frac{2e^\phi}{e^{2\phi}+1}\, d\phi,$$ which, to my understanding, can only be solved by setting $e^\phi=\tan\theta$. This substitution, however, is precisely identical to deriving $(1)$, setting $x=\tan\theta.$ My question is if it is possible to derive $(1)$ solely through hyperbolic substitutions---that is, devoid of trigonometric substitutions and knowledge of the result we are attempting to obtain?

Answer 1

You could always adapt the usual trick for secants, which gives gives

$$\begin{align*} \int \mathrm{sech}(x) \, \mathrm{d}x &= \int \frac{\mathbf{i} \mathrm{sech}(x)^2 + \mathrm{sech}(x) \mathrm{tanh}(x)}{\mathbf{i} \mathrm{sech}(x) + \mathrm{tanh}(x)} \, \mathrm{d}x \\&= \mathbf{i} \int \frac{\mathrm{d} \left(\mathbf{i} \mathrm{sech}(x) + \mathrm{tanh}(x) \right)}{\mathbf{i} \mathrm{sech}(x) + \mathrm{tanh}(x)} \\&= \mathbf{i} \log \left( \mathbf{i} \mathrm{sech}(x) + \mathrm{tanh}(x) \right) + C \end{align*}$$

although it's a bit of a mess if you want to turn it into $\arctan$ because you have to deal with its formula in terms of complex logarithms:

$$ \arctan(x) = \frac{1}{2 \mathbf{i}} \log\left( \frac{1 + \mathbf{i} x}{1 - \mathbf{i} x} \right) $$

You may be interested, however, in the fact that this integral is itself a special function that has some neat properties: the Gudermannian function.

Answer 2

You could use partial fractions $\int \frac 1{1+x^2} dx =\frac 12\int \frac i{x+i} -\frac i{x-i} dx \\ \frac i2 \ln \frac {x+i}{x-i} + C$

And now we would need to use Euler's identity to back into $\frac {x+i}{x-i}=e^{-2i\tan^{-1} x}$