What is the integral of (sech(1/x)tanh(1/x))/(x^2) ?

The integral of (sech(1/x)tanh(1/x))/(x^2) is sech(1/x)+C

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integral of (sech(1/x)tanh(1/x))/(x^2)

\int \frac{sech ( \frac{1}{x} ) tanh ( \frac{1}{x} )}{x ^{2}} d x

sech (\frac{1}{x}) + C

Solution steps

\int \frac{sech ( \frac{1}{x} ) tanh ( \frac{1}{x} )}{x ^{2}} d x

Apply u-substitution

= \int - sech (u) tanh (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int sech (u) tanh (u) d u

Apply u-substitution

= - \int - 1 d v

Integral of a constant:

\int a d x = a x

= - (- 1) v

Substitute back

= - (- 1) sech (\frac{1}{x})

Simplify

= sech (\frac{1}{x})

Add a constant to the solution

= sech (\frac{1}{x}) + C

What is the integral of (sech(1/x)tanh(1/x))/(x^2) ?
The integral of (sech(1/x)tanh(1/x))/(x^2) is sech(1/x)+C