Enraging Pythagoras 2.0

The question is called Enraging Pythagoras 2.0 because legend has it that the person who discovered the incommensurability (in modern terms, the irrationality) of $\sqrt{2}$ sent Pythagoras into such a rage that he had the "heretic" executed. The standard proof is a proof by contradiction, which I leave to the reader to discover themselves. We follow a similar proof technique here.

That the integral does not yield easily (or at all, for that matter) suggests that $f$ is indeed not rational. Since it is probably easier to proceed by assuming it is rational, we argue by contradiction. Assume that $\mathrm{P}(x)$ and $\mathrm{Q}(x)$ share no common factors, ie they are coprime.

$\frac{d}{dx} \left( \frac{\mathrm{P}}{\mathrm{Q}}e^x \right) = \frac{\mathrm{P}'\mathrm{Q} - \mathrm{PQ}' + \mathrm{PQ}}{\mathrm{Q}^2} \cdot e^x = \frac{e^x}{1+x^2}.$

Hence, $(\mathrm{P}'\mathrm{Q} - \mathrm{PQ}' + \mathrm{PQ})(1+x^2) = \mathrm{Q}^2$ . This implies that $\mathrm{Q}^2$ , and thus, $\mathrm{Q}$ , has a factor of $1+x^2$ .

Suppose that $\mathrm{Q} = (1+x^2)^n\mathrm{R}$ , such that $\mathrm{R}$ is not divisible by $1+x^2$ . Then $\dfrac{\mathrm{Q}^2}{1+x^2} = \mathrm{P}'\mathrm{Q} - \mathrm{PQ}' + \mathrm{PQ} = \mathrm{Q}(\mathrm{P}'+\mathrm{P}) - \mathrm{PQ}'$ has a factor of $(1+x^2)^{2n-1}$ , which implies that it has a factor of $(1+x^2)^n$ . This means that $\mathrm{PQ}' = \mathrm{P} \cdot (n(1+x^2)^{n-1}\mathrm{R} + (1+x^2)^n\mathrm{R}')$ has a factor of $(1+x^2)^n$ , so $\mathrm{PR}$ has a factor of $1+x^2$ . Since $\mathrm{R}$ cannot contribute that factor, it must come from $\mathrm{P}$ . However, this is a contradiction, since $\mathrm{P}$ was assumed to be coprime to $\mathrm{Q}$ . Therefore, we may conclude that $f$ is not a rational function, and the statement is false . $\square$

Before starting to prove it, let's state the Liouville's theorem:
The integral $I:=\int{f(x)e^{g(x)}}dx$ , for $f(x)$ and $g(x)$ rational functions, i.e. $f;g \in \mathbb{C}(X)$ is elementary iff there is $h \in \mathbb{C}(X)$ such that $f(x)=h'(x)+h(x)g(x)$ .
In our case, $f(x)=\frac{1}{1+x^2}$ and $g(x)=x$ . Let's now assume that $h$ is a function that satisfies the above conditions.
Hence we get the following differential equation: $\frac{1}{1+x^2}=h'(x)+xh(x)$ , which is of the form: $h'(x)+p(x)h(x)=q(x)$ , whose general solution is: $h(x)= \large \frac{\int{e^{\int{p(x)dx}}q(x)dx}+C}{e^{\int{p(x)dx}}}$ . Substituting with $f(x)=p(x)$ and $q(x)=g(x)$ , we get after simplifications(without caring about C) that $h(x)= \large \frac{\int {\frac{e^{\frac{x^{2}}{2}}}{1+x^{2}}dx}} {e^{\frac{x^{2}}{2}}}$ . If we substitute $x$ with $u:=1+x^{2}$ , we get that $\large \int {\frac{e^{\frac{x^{2}}{2}}}{1+x^{2}}dx}=e^{-\frac{1}{2}} \large \int{\frac{e^{\frac{u}{2}}}{u}dx}$ . To integrate the RHS of the last equation we will need to integrate $J:=\large \int{\frac{e^{\frac{u}{2}}}{u}du}$ , but this is just the well-known exponential integral $Ei(\frac{x}{2})=J$ and it was proven that this function is NOT elementary.
Hence, our function $h(x)$ is not elementary, therefore not a rational function. This contradicts our assumption that $h(x)$ is rational and hence our integral is not elementary.
Therefore, the statement is false.

This problem can be tackle simply as we can use substitution method ,that is let denominator be equal to T and then diff. Both sides put in required integral and then solving a simple integral in variable T we get something other than rational function ( logarithm function in x after subs. value of T in solved integral). Hence, false.