Solve the integral. \int_{0}^{1}\arcsin x dx

Factor each polynomial. If a polynomial cannot be factored, write prime. Factor out the greatest common factor as necessary:
${\displaystyle x^2+4x-5}$

The eccentricities of conic sections with one focus at the origin and the directrix corresponding and sketch a graph:
${\displaystyle e=2,}$
${\displaystyle r=-2\sec \theta .}$

The function:
${\displaystyle f\left(x\right)=e^{x+1}-2}$
a) Domain
b) Range
c) x, y-intercepts
d) Sketch a graph of f
e) What transformations happened to ${\displaystyle y=e^x}$ to get ${\displaystyle f\left(x\right)?}$

Evaluate the double integral ${\displaystyle x{y}^2}$ dA, D is enclosed by x=0 and ${\displaystyle x={(1-y^2)}^{\frac{1}{2}}}$

To calculate: The simplified form of the expression , ${\displaystyle \frac{12}{\sqrt{6}+\sqrt{2}}}$

Suppose I am given a Cauchy-Euler form second order differential equation
$x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y=f(x).$
The usual textbook method for solving the Cauchy Euler equation is to blackuce it to a linear differential equation with constant coefficients by the transformation $x=e^t$. But I have a fundamental doubt here, we know that $e^t>0$ $\mathrm{\forall }t\in \mathbb{R}$. But when we are using the above transformation we are subconsciously assuming $x>0$. How does this make sense?

Substituting $t=\ln (x)$ also makes no difference as ln is defined on ${\mathbb{R}}_{>0}$. So I now doubt the validity of the method followed to solve the Cauchy-Euler equation.

Can someone give me a proper explanation of what exactly going on here and why the process is valid?