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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Expand

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

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

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

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

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

Add a constant to the solution

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

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