What is the integral of (sec^2(8/x))/(x^2) ?

The integral of (sec^2(8/x))/(x^2) is -1/8 tan(8/x)+C

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integral of (sec^2(8/x))/(x^2)

\int \frac{sec ^{2} ( \frac{8}{x} )}{x ^{2}} d x

- \frac{1}{8} tan (\frac{8}{x}) + C

Solution steps

\int \frac{sec ^{2} ( \frac{8}{x} )}{x ^{2}} d x

Apply u-substitution

: \int - \frac{sec ^{2} ( u )}{8} d u

= \int - \frac{sec ^{2} ( u )}{8} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= - \frac{1}{8} tan (u)

Substitute back

u = \frac{8}{x}

= - \frac{1}{8} tan (\frac{8}{x})

Add a constant to the solution

= - \frac{1}{8} tan (\frac{8}{x}) + C

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What is the integral of (sec^2(8/x))/(x^2) ?
The integral of (sec^2(8/x))/(x^2) is -1/8 tan(8/x)+C