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$\int \frac{1+\tan x}{x+\log \sec x} d x=$
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1412 Upvotes
Verified Answer
The correct answer is:
$\log (x+\log \sec x)+c$
Put $t=x+\log \sec x \Rightarrow d t=(1+\tan x) d x$, then
$\begin{aligned}
\int \frac{1+\tan x}{x+\log \sec x} d x & =\int \frac{1}{t} d t=\log t+c \\
& =\log (x+\log \sec x)+c
\end{aligned}$
$\begin{aligned}
\int \frac{1+\tan x}{x+\log \sec x} d x & =\int \frac{1}{t} d t=\log t+c \\
& =\log (x+\log \sec x)+c
\end{aligned}$
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