(15) $\\int\\sec x\\tan xdx$\n(16) $\\int xe^{x}dx$

Explanation:

Step1: Solve $\int\sec x\tan xdx$

Recall the derivative formula: $(\sec x)'=\sec x\tan x$. By the fundamental theorem of calculus $\int f'(x)dx = f(x)+C$ (where $C$ is the constant of integration), so $\int\sec x\tan xdx=\sec x + C$.

Step2: Solve $\int xe^{x}dx$

Use integration - by - parts formula $\int u dv=uv-\int v du$. Let $u = x$, then $du=dx$; and $dv = e^{x}dx$, then $v = e^{x}$. Substitute into the formula: $\int xe^{x}dx=xe^{x}-\int e^{x}dx$. Since $\int e^{x}dx=e^{x}+C$, we have $\int xe^{x}dx=xe^{x}-e^{x}+C=(x - 1)e^{x}+C$.

Answer:

(15) $\sec x + C$; (16) $(x - 1)e^{x}+C$