Show that $\frac{3+x}{1+\cos^{2}x}$ is of bounded variation on $[0,2\pi]$. 
Show that $\frac{3+x}{1+\cos^{2}x}$ is of bounded variation on $[0,2\pi]$.

My idea is that we use the fact that this function is monotonic on subintervals of $[0,2\pi]$. Is this enough to show that though?
 A: Note that for all $x \in [0,2\pi]$
$$|f'(x)| = \left|\frac{(1+ \cos^2x) + (3+x)(2\cos x \sin x)}{(1+ \cos^2 x)^2} \right| \leqslant \frac{1 + 1+ (3 + 2\pi)(2)}{(1+0)^2} = 8 + 4\pi$$
For any partition, $P:0 = x_0 < x_1 < \ldots < x_n = 2\pi$, there exists by the mean value theorem $\xi_j \in (x_{j-1},x_j)$ such that
$$\begin{align}V(f;P) &= \sum_{j=1}^n|f(x_j) - f(x_{j-1})| \\&= \sum_{j=1}^n|f'(\xi_j)|(x_j - x_{j-1}) \\&\leqslant (8+4\pi)\sum_{j=1}^n (x_j - x_{j-1}) \\&= (8+4\pi)(2\pi)\end{align}$$
Hence, the total variation $V_0^{2\pi}(f) = \sup_P V(f;P) \leqslant 16\pi + 8\pi^2$
A: $$V^b_a(f)=\sup_{p\in P}\sum_{i=0}^{n_p-1}|f(x_{i+1})-f(x_i)|\\
\sup_{p\in P}\sum_{i=0}^{n_p-1}|\frac{3+x_{i+1}}{1+cos^{2}x_{i+1}}-\frac{3+x_{i}}{1+cos^{2}x_{i}}|$$now work with ,for the big enough $n$ we have $$x_{i+1}-x_i \to 0 \\so \\ x_{i+1} \sim x_i \to \cos(x_{i+1}) \sim \cos(x_{i})$$
$$|\frac{3+x_{i+1}}{1+cos^{2}x_{i+1}}-\frac{3+x_{i}}{1+cos^{2}x_{i}}|\\\leq 
|\frac{3+x_{i+1}}{1+\min\{cos^{2}x_{i+1},cos^{2}x_{i} \}}-\frac{3+x_{i}}{1+\min\{cos^{2}x_{i+1},cos^{2}x_{i} \}}|
\\\leq|\frac{3+x_{i+1}}{1+0}-\frac{3+x_{i}}{1+0}|=|x_{i+1}-x_{i}|$$
so
$$V^b_a(f)=\sup_{p\in P}\sum_{i=0}^{n_p-1}|f(x_{i+1})-f(x_i)|\leq \sup_{p\in P}\sum_{i=0}^{n_p-1}|x_{i+1}-f(x_i)|\leq 2\pi-0$$
Alternatively 
If f is differentiable and its derivative is Riemann-integrable, its total variation is the vertical component of the arc-length of its graph, that is to say,
$$V^b_a(f)=\int_{a}^{b}|f'(x)|dx$$ or proove $$\int_{a}^{b}\sqrt{1+(f'(x))^2}dx<\infty$$
