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

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

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

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

ln ∣ x ∣ - \frac{2}{x} + tanh (\frac{1}{x}) + C

Solution steps

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

Apply u-substitution

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int 1 d u = u

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

= - ln ∣ u ∣ - u - (u - tanh (u))

Substitute back

u = \frac{1}{x}

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

Simplify

- ln \frac{1}{x} - \frac{1}{x} - (\frac{1}{x} - tanh (\frac{1}{x})) : ln ∣ x ∣ - \frac{2}{x} + tanh (\frac{1}{x})

= ln ∣ x ∣ - \frac{2}{x} + tanh (\frac{1}{x})

Add a constant to the solution

= ln ∣ x ∣ - \frac{2}{x} + tanh (\frac{1}{x}) + C

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