∫{1+2tanx(tanx+secx)}12dx is equal to

∫{1+2tan2x+2tanxsecx}12 dx=∫(sec2x+tan2x+2tanx secx)12 dx=∫(secx+tanx) dx=logsecx(secx+tanx)+C

∫{1+2tanx(tanx+secx)}12dx is equal to

∫{1+2tan2x+2tanxsecx}12 dx=∫(sec2x+tan2x+2tanx secx)12 dx=∫(secx+tanx) dx=logsecx(secx+tanx)+C

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$\int {1 + 2 \tan x (\tan x + \sec x)}^{\frac{1}{2}} d x$ is equal to

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$\log | \sec x (\sec x - \tan x) | + C$

$\log | c o \sec x (\sec x + \tan x) | + C$

$\log | \sec x (\sec x + \tan x) | + C$

$\log | (\sec x + \tan x) | + C$

answer is C.

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Detailed Solution

$\int {1 + 2 \tan^{2} x + 2 \tan x \sec x}^{\frac{1}{2}} d x = \int {(\sec^{2} x + \tan^{2} x + 2 \tan x \sec x)}^{\frac{1}{2}} d x$

$= \int (\sec x + \tan x) d x = logsec x (\sec x + \tan x) + C$

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$\int {1 + 2 \tan x (\tan x + \sec x)}^{\frac{1}{2}} d x$ is equal to