How to calculate $\int_{0}^{4} \sqrt{\frac{t+4}{t}}dt$? I need help with this problem. I'm calculating the length of the smooth simple arc of the portion of the parabola $y^2=16x$ which lies between the lines $x=0$ and $x=4$. 
So I first parametrized the function like this: $t=x\Rightarrow y=\pm\sqrt{16t}$, thus $f(t)=(t, \pm\sqrt{16t})$. I first started with the positive square root. I calculated the derivative and the norm of the derivative: $f'(t)=(1,\frac{2}{\sqrt{t}})$ and $\Vert f'(t)\Vert=\sqrt{1^2+(\frac{2}{\sqrt{t}})^2}=\sqrt{\frac{t+4}{t}}$. After that, I used the formula of the length:$$l(t)=\int_{0}^{4} \sqrt{\frac{t+4}{t}}dt$$ the problem is that I don't know how to integrate that. Please help me.

Hint
$$\dfrac4t=\cot^2u\iff t=?,0\le u\le\dfrac\pi4$$
Use Indefinite integral of secant cubed
