Question

$$\int\frac{\sec^{2}x}{1+lan^{13}}\cdot Jor$$

Answer

$$int*sec(x)^2+Jo*l*a*n^13*r$$

Solution


Simplify  \(x\times 1\)  to  \(x\).
\[int\sec^{2}x+la{n}^{13}Jor\]
Regroup terms.
\[int\sec^{2}x+Jola{n}^{13}r\]