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.

You can also set $u = 1+\frac4t$ to obtain
$$\int_0^4 \sqrt{1+\frac4t}\,dt = \begin{bmatrix} u = 1+\frac4t \\ du = -4(u-1)^2\,dt \end{bmatrix} = \int_{2}^\infty \frac{\sqrt{u}}{4(u-1)^2}\,du$$
You can decompose the latter function as
$$\frac{\sqrt{u}}{4(u-1)^2} = \frac1{16}\left(\frac1{(\sqrt{u}-1)^2}-\frac1{(\sqrt{u}+1)^2}\right)$$
Then notice that 
$$\int \frac{du}{(\sqrt{u}\pm1)^2} = \begin{bmatrix} z = \sqrt{u}\pm 1 \\ du = 2(z\mp 1)\,dz \end{bmatrix} = \int \frac{2(z\mp 1)}{z^2}\,dz$$
which should be solvable.
