Calculating a line integral along a curve I am struggling with the following question;
how can I calculate the line integral $\int_C f(r) \, dr$, where $C$ starts at the point $(0,0,0)^\top$, and ends at the point $(a,a,a)^\top$ along the curve with $x=y=z=t$. The vector field is
$$
f(x,y,z) = \frac{-2}{x^2+y^2+z^2+a^2} \, (x,y,z)^\top.
$$
How do you solve this? ($(x,y,z)^\top$ being a column vector).
I did the following, but am not sure if it is right:

Thanks for any help I can get 

Please note that
$ ~ \vec F(x, y, z) = \frac{-2}{x^2+y^2+z^2+a^2} \, (x,y,z) ~$ is a conservative vector field because we have a potential function $g(x, y, z)$ such that $ \vec F(x, y, z) = \nabla g(x, y, z)$.
The potential function $g(x, y, z)$ is given by $g(x, y, z) = - \ln (x^2+y^2+z^2+a^2) \tag1$
By Fundamental Theorem of Line Integral,
$ \displaystyle \int_C \vec F \cdot dr = \int_C (\nabla g) \cdot dr = g(B) - g(A)$
where $A$ is the starting point and $B$ is the end point on the curve $C$.
In this case, $A = (0, 0, 0), B = (a, a, a)$. Plugging into $(1)$, the line integral is
$ - \ln (4a^2) + \ln (a^2) = - \ln 4$
