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Question

Question: \(\log(\sec^{2}x) + c\)...

log⁡(sec⁡2x)+c\log(\sec^{2}x) + c

A

log⁡(1+tan⁡x)+c\log(1 + \tan x) + c

B

−1(1+tan⁡x)2+c- \frac{1}{(1 + \tan x)^{2}} + c

C

∫e2x−1e2x+1 6mudx=\int_{}^{}\frac{e^{2x} - 1}{e^{2x} + 1}\mspace{6mu} dx =

D

e2x−1e2x+1+c\frac{e^{2x} - 1}{e^{2x} + 1} + c

Answer

log⁡(1+tan⁡x)+c\log(1 + \tan x) + c

Explanation

Solution

a=π4, 6mub=3a = \frac{\pi}{4},\mspace{6mu} b = 3

Put a=−π4, 6mub=3a = - \frac{\pi}{4},\mspace{6mu} b = 3 then it reduces to

a=π4, 6mub=a = \frac{\pi}{4},\mspace{6mu} b =.