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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= - tan (u)

Substitute back

u = \frac{1}{x}

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

Add a constant to the solution

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

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