How do I compute $\int_{0}^{2\pi} \frac{x(t)x'(t) + y(t)y'(t)}{x^2(t) + y^2(t)}dt$?

Question

I'm a little rusty on this stuff, so bear with me.

Let $F(t) = (x(t), y(t))$ be a closed continuously differentiable curve in $\mathbb{R}^2 \setminus \{(0,0) \}$. How do I compute the following integral?

$$\int_{0}^{2\pi} \frac{x(t)x'(t) + y(t)y'(t)}{x^2(t) + y^2(t)}dt .$$

I believe there's a substitution I can make. That is, if I let $u = \ln(x^2(t) + y^2(t))$, then the integral reduces to $$ \int_{t=0}^{t=2\pi}du, $$ which is rather easy to deal with. But, when I started to compute $du/dt$, I realized that I'm actually not quite sure how to take such a derivative. That is, we clearly have that $$\frac{du}{dt} = \frac{(x^2(t))' + (y^2(t))'}{x^2(t) + y^2(t)} ,$$ but what is, for example, $(x^2(t))'?$ By the chain rule, I assume it is something like $(x(x(t)))' = x'(x(t))x'(t)$. But that doesn't seem right... and isn't quite what I want, which would be $x(t)x'(t)$.

I feel like I'm perhaps over-looking something quite obvious...

Answer 1

$$x^2(t)=x(t)\cdot x(t)$$ so $$(x^2(t))'=2x(t)\cdot x'(t)$$ Your mistake was to think that $$x^2(t)=x(x(t))$$

Answer 2

$$\frac{du}{dt} = \frac{(x^2(t))' + (y^2(t))'}{x^2(t) + y^2(t)}=$$

$$\frac{2x(t)x'(t) + 2y(t)y'(t)}{x^2(t) + y^2(t)}$$