Evaluate the folowing path integrals $\int_{c}fds$ and the following are given:
$f(x,y,z) = \frac{x+y}{y+z}$ $\quad$ $c(t)=(t,\frac{2}{3}t^{3/2},t)$
I tried and then got stuck at this step. I'm not sure what trick I could use to continue from there.
$\int\frac{2t\sqrt{2+t}}{\frac{2}{3}t^{2/3}+t}dt$
= $\int\frac{2\sqrt{2+t}}{\frac{2}{3}\sqrt{t}+1}dt$
=6$\int\frac{\sqrt{2+t}}{2\sqrt{t}+3}dt$
A little hint will be really appreciated.
Since
$$c'(t)=\left(t,\,\frac23t^{3/2},\,t\right)\implies \left\|c'(t)\right\|=\sqrt{1+t+1}=\sqrt{t+2}$$
and our integral is
$$\int_0^1\frac{t+\frac23t^{3/2}}{\frac23t^{3/2}+t}\cdot\sqrt{t+2}\,dt=\int_0^1\sqrt{t+2}\,dt=\left.\frac23(t+2)^{3/2}\right|_0^1=$$
$$=\frac23\left(3^{3/2}-2^{3/2}\right)=2\sqrt3-\frac{4\sqrt2}3$$