\[\frac{\sec^{p + 1}x}{p + 1} + c] | MATH - INTEGRATION BY SUBSTITUTION | SolveeIt

$\frac{\sec^{p + 1}x}{p + 1} + c$

$\frac{\sec^{p}x}{p} + c$

$\frac{\tan^{p + 1}x}{p + 1} + c$

$\frac{\tan^{p}x}{p} + c$

None of these

Answer

$\frac{\tan^{p}x}{p} + c$

Explanation

$2(\sec x + \tan x) - x + c$

Put $1/3(\sec x + \tan x)^{3} + c$

then it reduces to

$\sec x(\sec x + \tan x) + c$ .

Question: \[\frac{\sec^{p + 1}x}{p + 1} + c\]...