What is the integral of (4e^{-3/(x^5)})/(x^6) ?

The integral of (4e^{-3/(x^5)})/(x^6) is 4/15 e^{-3/(x^5)}+C

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integral of (4e^{-3/(x^5)})/(x^6)

\int \frac{4 e ^{- \frac{3}{x ^{5}}}}{x ^{6}} d x

\frac{4}{15} e^{- \frac{3}{x ^{5}}} + C

Solution steps

\int \frac{4 e ^{- \frac{3}{x ^{5}}}}{x ^{6}} d x

Take the constant out:

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

= 4 \cdot \int \frac{e ^{- \frac{3}{x ^{5}}}}{x ^{6}} d x

Apply u-substitution

: \int \frac{e ^{- \frac{3}{u}}}{5 u ^{2}} d u

= 4 \cdot \int \frac{e ^{- \frac{3}{u}}}{5 u ^{2}} d u

Take the constant out:

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

= 4 \cdot \frac{1}{5} \cdot \int \frac{e ^{- \frac{3}{u}}}{u ^{2}} d u

Apply u-substitution

: \int \frac{e ^{v}}{3} d v

= 4 \cdot \frac{1}{5} \cdot \int \frac{e ^{v}}{3} d v

Take the constant out:

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

= 4 \cdot \frac{1}{5} \cdot \frac{1}{3} \cdot \int e^{v} d v

Use the common integral:

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

= 4 \cdot \frac{1}{5} \cdot \frac{1}{3} e^{v}

Substitute back

= 4 \cdot \frac{1}{5} \cdot \frac{1}{3} e^{- \frac{3}{x ^{5}}}

Simplify

4 \cdot \frac{1}{5} \cdot \frac{1}{3} e^{- \frac{3}{x ^{5}}} : \frac{4}{15} e^{- \frac{3}{x ^{5}}}

= \frac{4}{15} e^{- \frac{3}{x ^{5}}}

Add a constant to the solution

= \frac{4}{15} e^{- \frac{3}{x ^{5}}} + C

\frac{\partial}{\partial y} ((x + y) e^{y^{2}})

3 t + 4

f (x) = x^{3} + x^{2} + x + 1

\int e^{x} cos (e^{x}) d x

What is the integral of (4e^{-3/(x^5)})/(x^6) ?
The integral of (4e^{-3/(x^5)})/(x^6) is 4/15 e^{-3/(x^5)}+C