Chapter 15 - Section 15.3 - Path Independence, Conservative Fields, and Potential Functions - Exercises - Page 849: 11

$x \ln x-x+\tan (x+y)+\dfrac{\ln{(y^2+z^2)}}{2}+C$

Here, $f=[ \ln x +\sec^2 (x+y)]i +(\sec^2 (x+y) +\dfrac{y}{y^2+z^2}) j+(\dfrac{z}{y^2+z^2}) k$ This implies that $F=[x \ln x-x+\tan (x+y)]i +(\tan (x+y) +\dfrac{\ln{y^2+z^2}}{2}) j+(\dfrac{\ln{y^2+z^2}}{2}) k$ Hence, $F=x \ln x-x+\tan (x+y)+\dfrac{\ln{(y^2+z^2)}}{2}+C$