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$$
\int[1+2 \tan x(\tan x+\sec x)]^{\frac{1}{2}} d x=
$$
Options:
\int[1+2 \tan x(\tan x+\sec x)]^{\frac{1}{2}} d x=
$$
Solution:
2837 Upvotes
Verified Answer
The correct answer is:
$\log [\sec x(\sec x+\tan x)]+c$
Let $\left.I=\int[1+2 \tan x(\tan x+\sec x)]^{1 / 2} d x=\int 1+2 \tan ^2 x+2 \tan x \sec x\right)^{1 / 2} d x$
$$
\int\left[\left(1+\tan ^2 x\right)+\tan ^2 x+2 \sec x \tan x\right]^{1 / 2} d x=\int\left(\sec ^2 x+\tan ^2 x+2 \sec x \tan x\right)^{1 / 2} d x
$$
$$
\begin{aligned}
& =\int\left[(\sec x+\tan x)^2\right]^{1 / 2} d x=\int(\sec x+\tan x) d x=\int \sec x d x+\int \tan x d x \\
& =\log |\sec x+\tan x|-\log |\cos x|+c \quad=\log \frac{|\sec x+\tan x|}{|\cos x|}+c \\
& =\log [\sec x(\sec x+\tan x)]+c
\end{aligned}
$$
$$
\int\left[\left(1+\tan ^2 x\right)+\tan ^2 x+2 \sec x \tan x\right]^{1 / 2} d x=\int\left(\sec ^2 x+\tan ^2 x+2 \sec x \tan x\right)^{1 / 2} d x
$$
$$
\begin{aligned}
& =\int\left[(\sec x+\tan x)^2\right]^{1 / 2} d x=\int(\sec x+\tan x) d x=\int \sec x d x+\int \tan x d x \\
& =\log |\sec x+\tan x|-\log |\cos x|+c \quad=\log \frac{|\sec x+\tan x|}{|\cos x|}+c \\
& =\log [\sec x(\sec x+\tan x)]+c
\end{aligned}
$$
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