Let $\vec{F}=[(2yz^2+\ln x+\sec^2(x+y)]\,\hat{i}+[2xz^2-2ze^{yz}+\sec^2(x+y)+\frac{y}{y^2+z^2}]\,\hat{j}+[4xyz-2ye^{yz}+\frac{z}{y^2+z^2}]\,\hat{k}$
How to find all the function of $f$ such that $\vec{\nabla}f=\vec{F}$?
I tried to integrate each component of $\vec{F}$, is this method correct?
Can anyone help me? Thank you so much.
$$\vec{F}=[(2yz^2+\ln x+\sec^2(x+y)]\,\hat{i}+[2xz^2-2ze^{yz}+\sec^2(x+y)+\frac{y}{y^2+z^2}]\,\hat{j}+[4xyz-2ye^{yz}+\frac{z}{y^2+z^2}]\,\hat{k}$$ First integration gives: $$I_1=\int [(2yz^2+\ln x+\sec^2(x+y)]dx$$ $$I_1=2yz^2x+x \ln x - x + \tan (x+y)+ C_1$$ Where $c_1$ is all the functions that dosen't depend on $x$.
The second integral: $$I_2=\int [2xz^2-2ze^{yz}+\sec^2(x+y)+\frac{y}{y^2+z^2}] dy$$ $$I_2=2xz^2y-2e^{yz}+\tan(x+y)+\frac 12 \ln ({y^2+z^2}) +C_2$$ Last integral gives: $$I_3=\int [4xyz-2ye^{yz}+\frac{z}{y^2+z^2}]\, dz$$ $$I_3=2xyz^2-2e^{yz}+\frac 12\ln ({y^2+z^2})+C_3$$
Now you can deduce the function $f$: $$f(x,y,z)=2xz^2y-2e^{yz}+\tan(x+y)+\frac 12 \ln ({y^2+z^2}) +x\ln x - x +C$$