What is the integral of sinh(1/x) ?

The integral of sinh(1/x) is 1/2 (-e^{-1/x}x-\Ei(-1/x)+e^{1/x}x-\Ei(1/x))+C

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integral of sinh(1/x)

\int sinh (\frac{1}{x}) d x

\frac{1}{2} (- e^{- \frac{1}{x}} x - Ei (- \frac{1}{x}) + e^{\frac{1}{x}} x - Ei (\frac{1}{x})) + C

Solution steps

\int sinh (\frac{1}{x}) d x

Rewrite using trig identities

= \int \frac{- e ^{- \frac{1}{x}} + e ^{\frac{1}{x}}}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int - e^{- \frac{1}{x}} + e^{\frac{1}{x}} d x

Apply the Sum Rule:

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

= \frac{1}{2} (- \int e^{- \frac{1}{x}} d x + \int e^{\frac{1}{x}} d x)

\int e^{- \frac{1}{x}} d x = e^{- \frac{1}{x}} x + Ei (- \frac{1}{x})

\int e^{\frac{1}{x}} d x = e^{\frac{1}{x}} x - Ei (\frac{1}{x})

= \frac{1}{2} (- (e^{- \frac{1}{x}} x + Ei (- \frac{1}{x})) + e^{\frac{1}{x}} x - Ei (\frac{1}{x}))

Simplify

\frac{1}{2} (- (e^{- \frac{1}{x}} x + Ei (- \frac{1}{x})) + e^{\frac{1}{x}} x - Ei (\frac{1}{x})) : \frac{1}{2} (- e^{- \frac{1}{x}} x - Ei (- \frac{1}{x}) + e^{\frac{1}{x}} x - Ei (\frac{1}{x}))

= \frac{1}{2} (- e^{- \frac{1}{x}} x - Ei (- \frac{1}{x}) + e^{\frac{1}{x}} x - Ei (\frac{1}{x}))

Add a constant to the solution

= \frac{1}{2} (- e^{- \frac{1}{x}} x - Ei (- \frac{1}{x}) + e^{\frac{1}{x}} x - Ei (\frac{1}{x})) + C

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What is the integral of sinh(1/x) ?
The integral of sinh(1/x) is 1/2 (-e^{-1/x}x-\Ei(-1/x)+e^{1/x}x-\Ei(1/x))+C