Question

integral sec square (log x).dx upon x

Answer 1

I hope this answer helps you

Answer 2

$\int {\dfrac{\sec^2(\log x)}{x} } \, dx=\tan (\log x)+C$

Step-by-step explanation:

Let $I=\int {\dfrac{\sec^2(\log x)}{x} } \, dx$     ...(1)

To find, $\int {\dfrac{\sec^2(\log x)}{x} } \, dx=?$

Let $t=\log x$

$dt=\dfrac{1}{x} dx$

Now, (1) becomes

$I=\int\sec^2 t \, dt$

$=\tan t+C$    ....(2)

Where, C is called integration constant

[ ∵$\int\sec^2 x \, dx=\tan x+C$]

Put $t=\log x$ in (2), we get

$I=\tan (\log x)+C$

Hence, $\int {\dfrac{\sec^2(\log x)}{x} } \, dx=\tan (\log x)+C$