Try using complex numbers

$\begin{aligned} \int \frac{1}{1+x^2} dx & = \int \frac{\sec^2 \theta}{1+\tan^2 \theta} d \theta \quad \quad \small \color{#3D99F6}{\text{Let } x = \tan \theta \quad \Rightarrow dx = sec^2 \theta \space d \theta} \\ & = \int \frac{\sec^2 \theta}{\sec^2 \theta} d \theta = \int d \theta = \theta + C = \boxed{\arctan x} + C \end{aligned}$

changing x to z and factoring to get $1+z^2=(z+i)(z-i)$ using partial fractions we have $\Large\frac{1}{(z+i)(z-i)}=\frac{A}{z+i}+\frac{B}{z-i}=\frac{A(z+i)+B(z-i)}{z^2+1}$

if we let $z=i$ we'll get $A=\frac{i}{2}$ and $B=\frac{-i}{2}$ we now have $\displaystyle\int\frac{1}{1+x^2}dx=\frac{i}{2}\displaystyle\int(\frac{1}{z+i}-\frac{1}{z-i})dz$ this reduces to

$\frac{i}{2}(ln(\frac{z+i}{z-i}))=\frac{i}{2}(ln(\left|\frac{z+i}{z-i}\right|)+i arg(\frac{z+i}{z-i})$ ,where arg denotes the argument

if we let z be a real number equal to x we have $\left| x+i \right| =\left| x-i \right|$ so $ln(\frac{x+i}{x-i})=0$ this will also make the argument of $\frac{x+i}{x-i}$ equal to 2 $\theta$

simplifying we get $\displaystyle\int\frac{1}{1+x^2}dx =-\theta +C$

the trick here is to let $C=a+\frac{\pi}{2}$

if we let x be a side on a right triangle with height 1 and an angle $\theta$ ,we'll have $\frac{\pi}{2}-\theta =\arctan(x)$

so

$\displaystyle\int\frac{1}{1+x^2}dx=arctan(\theta)+a$

my solution doesn't include euler's formula in it, but i'm almost certainly sure there exists a solution which uses it

kudos to the one who'll publish it :)

From this graph, we find that the answer is arctan (x) + C (where C is the arbitrary constant of integration)
As, it is mentioned that it has to be ignored the final answer is $arctan (x)$ . $_\square$