I need to calculate the constant alpha so that the integral be independent of the path we take.
$$\int_C [x \ln{(x² + 1)}\;+\;(x² + 1)y]dx + \alpha[\frac{x³}{3} + x + \sin{y}]dy$$
Answer: $\alpha = 1$
My solution:
To find the value of alpha, We need, first, define the vectorial field as a simply connected (has no singularities). If we look at the field, it has nothing that restrict the domain of the field. So, it can be considered as a simply connected field.
My doubt: I got stuck in the next step. I don't know if I need to find a potential function for the vectorial field to determine $\alpha$ or if I need to take another path to calculate this.
A differential, $\alpha(x,y)dx+ \beta(x,y)dy$ is independent of the path if and only if it is an "exact differential"- that there is a function, f such that $df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial} dy= \alpha(x,y)dx+ \beta(x,y)dy$ which means that $\alpha(x,y)= \frac{\partial f}{\partial x}$ and $\beta(x,y)= \frac{\partial f}{\partial y}$. Then $\int_a^b df= f(b)- f(a)$ no matter what the path from a to b is.
If we differentiate $\frac{\partial f}{\partial x}$ with respect to y we get $\frac{\partial^2 f}{\partial x\partial y}= \frac{\partial \alpha(x,y)}{\partial y}$. If we differentiate $\frac{\partial f}{\partial y}$ with respect to x we get $\frac{\partial^2 f}{\partial y\partial x}= \frac{\partial \beta(x,y)}{\partial x}$.
For any "reasonably nice" functions that "mixed" deriatives are equal so we get the "cross condition" $\frac{\partial \alpha}{\partial y}= \frac{\partial \beta}{\partial x}$.
Here, that means that we must have $x^2+ 1= \alpha(x^2+ 1)$ so $\alpha= 1$.