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

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

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

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

- \frac{1}{x} + tanh (\frac{1}{x}) + C

Solution steps

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

Apply u-substitution

: \int - tanh^{2} (u) d u

= \int - tanh^{2} (u) d u

Take the constant out:

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

= - \int tanh^{2} (u) d u

Rewrite using trig identities

= - \int 1 - sech^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int 1 d u - \int sech^{2} (u) d u)

\int 1 d u = u

\int sech^{2} (u) d u = tanh (u)

= - (u - tanh (u))

Substitute back

u = \frac{1}{x}

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

Simplify

= - \frac{1}{x} + tanh (\frac{1}{x})

Add a constant to the solution

= - \frac{1}{x} + tanh (\frac{1}{x}) + C

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