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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - sec (u) d u

= \int - sec (u) d u

Take the constant out:

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

= - \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= - ln ∣ tan (u) + sec (u) ∣

Substitute back

u = \frac{1}{x}

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

Add a constant to the solution

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

\frac{d y}{d x} = (\frac{x - y}{x})

(cos (5 x))^{^{'}}

f (x) = \frac{6 x}{x + 5}

8 x e^{- k x}

n = 1 \sum \infty \frac{2 ^{n}}{( n ) !}

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