What is the integral of (e^{1/(x^4)})/(x^5) ?

The integral of (e^{1/(x^4)})/(x^5) is -1/4 e^{1/(x^4)}+C

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integral of (e^{1/(x^4)})/(x^5)

\int \frac{e ^{\frac{1}{x ^{4}}}}{x ^{5}} d x

- \frac{1}{4} e^{\frac{1}{x ^{4}}} + C

Solution steps

\int \frac{e ^{\frac{1}{x ^{4}}}}{x ^{5}} d x

Apply u-substitution

: \int \frac{e ^{\frac{1}{u}}}{4 u ^{2}} d u

= \int \frac{e ^{\frac{1}{u}}}{4 u ^{2}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{e ^{\frac{1}{u}}}{u ^{2}} d u

Apply u-substitution

: \int - e^{v} d v

= \frac{1}{4} \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}{4} (- \int e^{v} d v)

Use the common integral:

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

= \frac{1}{4} (- e^{v})

Substitute back

= \frac{1}{4} (- e^{\frac{1}{x ^{4}}})

Simplify

= - \frac{1}{4} e^{\frac{1}{x ^{4}}}

Add a constant to the solution

= - \frac{1}{4} e^{\frac{1}{x ^{4}}} + C

x \to 2 lim (\frac{4 x - 3}{x + 2})

(\frac{x ^{4}}{4})^{^{'}}

\frac{360}{2 s ^{2} + 10 s + 8}

\frac{\partial}{\partial x} (y e^{3 x} \cdot 3)

y = \frac{7}{x ^{2} + 3}, (2, 1)

What is the integral of (e^{1/(x^4)})/(x^5) ?
The integral of (e^{1/(x^4)})/(x^5) is -1/4 e^{1/(x^4)}+C