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

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

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

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

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

Solution steps

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

Simplify

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

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 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= - \frac{1}{2} (- u + tan (u))

Substitute back

u = \frac{2}{x}

= - \frac{1}{2} (- \frac{2}{x} + tan (\frac{2}{x}))

Simplify

- \frac{1}{2} (- \frac{2}{x} + tan (\frac{2}{x})) : \frac{1}{x} - \frac{1}{2} tan (\frac{2}{x})

= \frac{1}{x} - \frac{1}{2} tan (\frac{2}{x})

Add a constant to the solution

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

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