What is the integral of (sec^2(1/(x^3)))/(x^4) ?

The integral of (sec^2(1/(x^3)))/(x^4) is -1/3 tan(1/(x^3))+C

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integral of (sec^2(1/(x^3)))/(x^4)

\int \frac{sec ^{2} ( \frac{1}{x ^{3}} )}{x ^{4}} d x

- \frac{1}{3} tan (\frac{1}{x ^{3}}) + C

Solution steps

\int \frac{sec ^{2} ( \frac{1}{x ^{3}} )}{x ^{4}} d x

Apply u-substitution

: \int \frac{sec ^{2} ( \frac{1}{u} )}{3 u ^{2}} d u

= \int \frac{sec ^{2} ( \frac{1}{u} )}{3 u ^{2}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{sec ^{2} ( \frac{1}{u} )}{u ^{2}} d u

Apply u-substitution

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

= \frac{1}{3} \cdot \int - sec^{2} (v) d v

Take the constant out:

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

= \frac{1}{3} (- \int sec^{2} (v) d v)

Use the common integral:

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

= \frac{1}{3} (- tan (v))

Substitute back

= \frac{1}{3} (- tan (\frac{1}{x ^{3}}))

Simplify

= - \frac{1}{3} tan (\frac{1}{x ^{3}})

Add a constant to the solution

= - \frac{1}{3} tan (\frac{1}{x ^{3}}) + C

\int \frac{cos ( \frac{π}{x ^{19}} )}{x ^{20}} d x

\frac{d y}{d x} = \frac{1}{x} (y^{2} + 0.8 y)

\int (∣ x ∣) d x

\frac{d}{d x} (4 x^{2} + 3)

\int x 4 + x^{2} d x

What is the integral of (sec^2(1/(x^3)))/(x^4) ?
The integral of (sec^2(1/(x^3)))/(x^4) is -1/3 tan(1/(x^3))+C