Answer
$\displaystyle \sqrt{3}\ln(\frac{b}{a})$
Work Step by Step
We calculate the line integral as follows:
$\vec{r}(t) = \langle t, t, t\rangle\qquad\qquad\therefore\vec{r}'(t) =\langle 1, 1, 1\rangle$
$\displaystyle \int_a^bf(\vec{r}(t))|\vec{r}'(t)|dt = \int_a^b\frac{t+t+t}{t^2+t^2+t^2}\sqrt{1^2+1^2+1^2}dt$
$\displaystyle \sqrt{3}\int_a^b\frac{3t}{3t^2}dt=\sqrt{3}\int_a^b\frac{1}{t}dt$
$\displaystyle \sqrt{3}[\ln(t)]^{t = b}_{t = a}$
$\sqrt{3}[\ln(b) - \ln(a)]$
$\displaystyle \sqrt{3}\ln(\frac{b}{a})$
