What is wrong with this attempt to solve $x^x = 2$? For $x^x= 2$, hence the solution is :
$x\cdot \ln(x) = \ln(2)$
$x\cdot\int(1/x)\,dx = \ln(2)$
$\int(x/x)\,dx = \ln(2)$
$x+c = \ln(2)$, where $c = 0$
so, $x = \ln (2)$
Can someone tell me why this solution is wrong?
 A: I believe that the error is that you incorrectly used the linearity of integration. You couldn’t move the x inside the integral of 1/x, because x isn’t a constant, it is the variable you are integrating with respect to.
A: It should start:
$x^x = 2$
$x\cdot \ln(x) = \ln(2)$
$x\cdot\int_1^x(1/t)\,dt = \ln(2)$
And then you're stuck! Sure you can put the $x$ inside the integral, but you're integrating FROM 1 to $x$ with respect to a placeholder variable (I've used $t$). And in fact, it's basically impossible to get unstuck without "cheating". And by cheating, I mean using the Lambert W function...
The Lambert W function often shows up in solving these types of problems. We can just define $W(x)$ to be the inverse of $xe^x$, and this turns out to be something for which numerical approximations are well-known. This video explores how to solve the problem using that at a relatively accessible level: https://www.youtube.com/watch?v=WWyMRmV1hLg .
A: The error was in the following step
$$\int \frac{x}{x} dx$$
As you cannot transfer the $x$ into the integral that way because $x$ is not a constant. To bring it inside the integral, the new integral must evaluate to $x\ln x$. So the correct way to bring it inside will be
$$x\int\frac{1}{x}dx=\int(\ln x+1)dx$$
The actual solution requires the use of a special function called the lambert $W$ function where it has the property
$$W(xe^x)=x$$
To begin with, you need to transform your equation into the form $ue^u$, which is easy enough
$$x^x=2$$
$x=e^u$
$$e^{ue^u}=2$$
$$ue^u=\ln2$$
Using the lambert W function
$$u=W(\ln2)$$
Substituting back in for $x$
$$\ln(x)=W(\ln2)$$
$$x=e^{W(\ln2)}$$
There are many solutions for the lambert W function, however the only one with a real answer is $W(\ln 2) \approx 0.444436091$ which makes the final answer
$$x \approx 1.55961046946236$$
