## Thomas' Calculus 13th Edition

$$F=x \ln x-x+\tan (x+y)+ \dfrac{\ln{(y^2+z^2)}}{2}+C$$
We have $$F=[ \ln x +\sec^2 \space (x+y)]i +(\sec^2\space (x+y) +\dfrac{y}{y^2+z^2}) j+(\dfrac{z}{y^2+z^2}) \space k \\ \implies F=[x \ln x-x+\tan (x+y)]i +(\tan (x+y) +\dfrac{\ln{y^2+z^2}}{2}) \space j+(\dfrac{\ln{y^2+z^2}}{2}) \space k \\ =x \ln x-x+\tan (x+y)+ \dfrac{\ln{(y^2+z^2)}}{2}+C$$