What is the integral of (e^{1/(x^3)})/(x^4) ?

The integral of (e^{1/(x^3)})/(x^4) is -1/3 e^{1/(x^3)}+C

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integral of (e^{1/(x^3)})/(x^4)

\int \frac{e ^{\frac{1}{x ^{3}}}}{x ^{4}} d x

- \frac{1}{3} e^{\frac{1}{x ^{3}}} + C

Solution steps

\int \frac{e ^{\frac{1}{x ^{3}}}}{x ^{4}} d x

Apply u-substitution

: \int \frac{e ^{\frac{1}{u}}}{3 u ^{2}} d u

= \int \frac{e ^{\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{e ^{\frac{1}{u}}}{u ^{2}} d u

Apply u-substitution

: \int - e^{v} d v

= \frac{1}{3} \cdot \int - e^{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 e^{v} d v)

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{3} (- e^{v})

Substitute back

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

Simplify

= - \frac{1}{3} e^{\frac{1}{x ^{3}}}

Add a constant to the solution

= - \frac{1}{3} e^{\frac{1}{x ^{3}}} + C

\frac{\partial}{\partial x} (8 x^{2} cos (5) x^{3})

\frac{d}{d x} (\frac{ln ( x )}{e ^{x}})

\int_{0}^{2} \frac{1}{9 - x ^{2}} d x

f (x) = 3 x^{4}

x \to 7 lim (\frac{x ^{2} - 4}{7 - x})

What is the integral of (e^{1/(x^3)})/(x^4) ?
The integral of (e^{1/(x^3)})/(x^4) is -1/3 e^{1/(x^3)}+C