Arc length of curve stuck with integration Data from exercise
$$y=\frac{4}{3}x^2+2\\
x\in[-1,1]$$
Formula for length of curve
$$L=\int_a^{b}\sqrt{1+(f(x)')^2}\ dx$$
So far i have
$$y'=\frac{8}{3}x$$
$$\int_{-1}^{1}\sqrt{1+\frac{64}{9}x^2}\ dx$$
Substition
$$t^2=\frac{64}{9}x^2$$
$$t=\frac{8}{3}x$$
$$\frac{3}{8}dt=dx$$
So i have
$$\int_{-\frac{8}{3}}^{\frac{8}{3}}\sqrt{1+t^2}\ dt$$
I this point i dont have a clue how to integrate this

Let $x = \sinh \theta \implies$
$$
\begin{align}
I &= \int \sqrt{1+x^2}dx \\
&= \int \sqrt{1+\sinh^2 \theta}\cosh \theta d \theta \\
&= \int \cosh^2 \theta d \theta \\
&= \frac{1}{2}\int 1 +  \cosh 2 \theta d \theta \\
&= \frac{1}{2}\theta+\frac{1}{4}\sinh 2 \theta + c\\
&= \frac{1}{2}\sinh^{-1}x+\frac{1}{2}x\sqrt{1+x^2}+c\\
&=\frac{1}{2}\ln(x + \sqrt{1+x^2})+\frac{1}{2}x\sqrt{1+x^2}+c
\end{align}$$
