Tengo una duda acerca de cómo resolver este problema matemático

$$\begin{align}&L=\int_{-1}^2 \sqrt{1+\left(\dfrac{d}{dx}x^2\right)^2}~dx\\&\\&L=\int_{-1}^2\sqrt{1+4x^2}~dx\\&\\&\text{Cambio de variable para }\int \sqrt{1+4x^2}~dx\\&2x=\sinh u\to 2dx=\cosh u ~du\\&u=\ln|2x+\sqrt{1+4x^2}|\\&\\&L(u)=\dfrac{1}{2}\int \cosh^2 u ~du\\&\\&L(u)=\dfrac{1}{2}\int \cosh u ~d(\sinh u) \\&\\&L(u)=\dfrac{1}{2}\cosh u\sinh u-\dfrac{1}{2}\int\sinh u~ d(\cosh u)\\&\\&L(u)=\dfrac{1}{2}\cosh u\sinh u-\dfrac{1}{2}\int\sinh^2 u ~du\\&\\&L(u)=\dfrac{1}{2}\cosh u\sinh u-\dfrac{1}{2}\int\sinh^2 u ~du\\&\\&L(u)=\dfrac{1}{2}\cosh u\sinh u-\dfrac{1}{2}\int\cosh^2 u-1 ~du\\&\\&L(u)=\dfrac{1}{2}\cosh u\sinh u-\underbrace{\dfrac{1}{2}\int\cosh^2 u ~du}_{=L(u)}+\dfrac{1}{2}u\\&\\&L(u)=\dfrac{1}{4}\cosh u\sinh u+\dfrac{1}{4}u+C\\&\\&L(x)=\dfrac{1}{4}\sqrt{1+4x^2}\cdot (2x)+\dfrac{1}{4}\ln\left|2x+\sqrt{1+4x^2}\right|+C\\&\\&L(x)=\dfrac{1}{2}x\sqrt{1+4x^2}+\dfrac{1}{4}\ln\left|2x+\sqrt{1+4x^2}\right|+C\\&------\\&L=L(2)-L(-1)\\&\\&L=\sqrt{17}+\dfrac{1}{4}\ln(4+\sqrt{17})+\dfrac{1}{2}\sqrt{5}-\dfrac{1}{4}\ln(-2+\sqrt{5})\\&\\&L=\dfrac{1}{4}\ln\dfrac{\sqrt{17}+4}{\sqrt{5}-2}+\sqrt{17}+\dfrac{1}{2}\sqrt{5}\\&\\&\boxed{L=\dfrac{1}{4}\ln(\sqrt{35}+4\sqrt{5}+2\sqrt{17}+8)+\sqrt{17}+\dfrac{1}{2}\sqrt{5}}\end{align}$$

Se usó a las funciones trigonométricas hiperbólicas