<s>[INST] Compute $\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}$ Compute the given line integral:

*

*$$\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}$$

Let $P=\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}},Q=\frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}}, R=\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}$ then $$\color{blue}{\frac{\partial P}{\partial y}}=\frac{-xy}{(x^2+y^2+z^2)^{3/2}},\color{green}{\frac{\partial P}{\partial z}}=\frac{-xz}{(x^2+y^2+z^2)^{3/2}}$$$$\color{blue}{\frac{\partial Q}{\partial x}}=\frac{-xy}{(x^2+y^2+z^2)^{3/2}},\color{red}{\frac{\partial Q}{\partial z}}=\frac{-yz}{(x^2+y^2+z^2)^{3/2}}$$$$\color{green}{\frac{\partial R}{\partial x}}=\frac{-xz}{(x^2+y^2+z^2)^{3/2}},\color{red}{\frac{\partial R}{\partial y}}=\frac{-yz}{(x^2+y^2+z^2)^{3/2}}$$
So the vector field is conservative which means there is $f$ such that $\vec \nabla f=F$, I don't know how to continue.
 [/INST] If you have checked the completeness then your expression is definitely of the form
$$\frac{\partial U}{\partial x}dx + \frac{\partial U}{\partial y}dy + \frac{\partial U}{\partial z}dz$$
for some (potential) function $U$. Hence
$$\begin{cases}
\dfrac{\partial U}{\partial x} = \dfrac{x}{\sqrt{x^2+y^2+z^2}} \\ 
\dfrac{\partial U}{\partial y} = \dfrac{y}{\sqrt{x^2+y^2+z^2}} \\
\dfrac{\partial U}{\partial z} = \dfrac{z}{\sqrt{x^2+y^2+z^2}}
\end{cases}$$
Integrating the first equality w.r.t $x$ we have
$$\int \dfrac{\partial U}{\partial x}dx = \int\dfrac{x}{\sqrt{x^2+y^2+z^2}}dx$$
substitute $t(x) = \sqrt{x^2+y^2+z^2}$ in the integral on the RHS. Then, $$dt = \dfrac{x}{\sqrt{x^2+y^2+z^2}}dx$$ and therefore
$$\begin{align}
U &= \int dt \\
&= t + C(y,z) \\
&= \sqrt{x^2+y^2+z^2} + C(y,z) \tag{1}\end{align}$$
Similarly,
$$U = \sqrt{x^2+y^2+z^2} + C(x,y) \tag{2}$$
$$U = \sqrt{x^2+y^2+z^2} + C(z,x) \tag{3}$$
Now, you show that $C(x,y) = \text{const}$.
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