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

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

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

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

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

Solution steps

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

Simplify

\frac{1}{x ^{2}} sinh (\frac{1}{x}) : \frac{sinh ( \frac{1}{x} )}{x ^{2}}

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

Apply u-substitution

= \int - sinh (u) d u

Take the constant out:

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

= - \int sinh (u) d u

Use the common integral:

\int sinh (u) d u = cosh (u)

= - cosh (u)

Substitute back

u = \frac{1}{x}

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

Add a constant to the solution

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

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