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:

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

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.

Consider
$$f(x,y,z)=\sqrt{x^2+y^2+z^2}+C$$
then $\vec \nabla f=F$ and
$$\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}=f(3,4,5)-f(0,0,0)=5\sqrt{2}.$$
