Un ejercicio de integración con intervención de la función valor absoluto

Deseamos calcular el valor de la integral definida $$\displaystyle \int_{1}^{3}\,\left(\dfrac{x}{|x|}+\dfrac{2\,|x|}{x}\right)\,dx$$

Recordemos la definición de valor absoluto de un número real: $$\|x|:=\left\{\begin{matrix}x & \text{si} & x\gt 0 \\ -x & \text{si} & x \lt 0 \end{matrix}\right.$$

Notemos que $\forall x \in \mathbb{R}: x^2=\left(|x|\right)^2 \Leftrightarrow x \cdot x=|x|\cdot|x|\; \Leftrightarrow \dfrac{x}{|x|}=\dfrac{|x|}{x} \therefore \dfrac{x}{|x|}+2\cdot \dfrac{|x|}{x}=\dfrac{x}{|x|}\,\left( 1+2 \right)=\dfrac{3x}{|x|} $

Entonces, $\displaystyle \int_{1}^{3}\,\left(\dfrac{x}{|x|}+\dfrac{2\,|x|}{x}\right)\,dx=\int_{1}^{3}\,\dfrac{3x}{|x|}\,dx=3\,\int_{1}^{3}\,\dfrac{x}{|x|}\,dx = 3\,\int_{1}^{3}\,\dfrac{x}{x}\,dx \because\, |x|=x,\; \forall x \in [1,3]\subset \mathbb{R}$. En consecuencia, $$\displaystyle \int_{1}^{3}\,\left(\dfrac{x}{|x|}+\dfrac{2\,|x|}{x}\right)\,dx=3\,\int_{1}^{3}\,dx=\displaystyle 3\,\left[x\right]_{1}^{3}=3\cdot (3-1)=3\cdot 2=6$$ $\diamond$