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

Call the indefinite integral $I$
$$I\equiv\int\sqrt{1+t^2}\,\mathrm dt$$
Through integration by parts with $u=\sqrt{1+t^2}$ and $v=x$, then
$$I=t\sqrt{1+t^2}-\int\frac {t^2}{\sqrt{1+t^2}}\,\mathrm dt$$
The integrand can also be rewritten as
\begin{align*}
I & =\int\frac {1+t^2}{\sqrt{1+t^2}}\,\mathrm dt\\ & =\int\frac {\mathrm dt}{\sqrt{1+t^2}}+\int\frac {t^2}{\sqrt{1+t^2}}\,\mathrm dt
\end{align*}
Adding the two expressions for $I$ together and dividing both sides by two eliminates one of the integral terms.
\begin{align*}
I & =\frac 12t\sqrt{1+t^2}+\frac 12\int\frac {\mathrm dt}{\sqrt{1+t^2}}
\end{align*}
The last integral is somewhat well-known and can be easily evaluated with an Euler Substitution $x=t+\sqrt{1+t^2}$.
$$\int\frac {\mathrm dt}{\sqrt{1+t^2}}=\log\left(t+\sqrt{1+t^2}\right)$$
Therefore, the indefinite version of the integral becomes
$$\int\sqrt{1+t^2}\,\mathrm dt\color{blue}{=\frac 12t\sqrt{1+t^2}+\frac 12\log\left(t+\sqrt{1+t^2}\right)+C}$$
Now, substitute in the bounds for $t=8/3$ and $t=-8/3$ to arrive at your answer.
