The answer is: \frac{56\sqrt{77}}{3}\approx 163.8
First, parameterize the line segment. The quickest way to do this is to let P=(0,6,-1) and Q=(4,1,5) and, thinking of these as vectors, find a vector going from P to Q by subtracting the components/coordinates of P from the corresponding components/coordinates of Q to get \vec{v}=(4,-5,6). Now let \vec(c)(t)=P+t\vec{v}=(4t,6-5t,-1+6t) for 0\leq t\leq 1.
The velocity vector is \vec{c}'(t)=\vec{v}, and its length (the speed) is ||\vec{v}||=\sqrt{16+25+36}=\sqrt{77}. Letting f(x,y,z)=x^2z, the function we have to integrate as t goes from t=0 to t=1 is f(4t,6-5t,-1+6t)||\vec{v}||=\sqrt(77)(4t)^2\cdot (-1+6t)=\sqrt{77}(96t^3-16t^2).
(Note that ds=\frac{ds}{dt}dt=||\vec{v}||dt.)
Here's the integral calculation:
\int_{0}^{1}\sqrt{77}(96t^3-16t^2)dt=\sqrt(77)(24t^{4}-\frac{16}{3}t^{3})\|_{t=0}^{t=1}
=\sqrt{77}\cdot \frac{72-16}{3}=\frac{56\sqrt{77}}{3}\approx 163.8
