What is the integral of (2e^{-6/(x^4)})/(x^5) ?

The integral of (2e^{-6/(x^4)})/(x^5) is 1/12 e^{-6/(x^4)}+C

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integral of (2e^{-6/(x^4)})/(x^5)

\int \frac{2 e ^{- \frac{6}{x ^{4}}}}{x ^{5}} d x

\frac{1}{12} e^{- \frac{6}{x ^{4}}} + C

Solution steps

\int \frac{2 e ^{- \frac{6}{x ^{4}}}}{x ^{5}} d x

Take the constant out:

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

= 2 \cdot \int \frac{e ^{- \frac{6}{x ^{4}}}}{x ^{5}} d x

\frac{e ^{- \frac{6}{x ^{4}}}}{x ^{5}} = e^{- \frac{6}{x ^{4}}} \frac{1}{x ^{5}}

= 2 \cdot \int e^{- \frac{6}{x ^{4}}} \frac{1}{x ^{5}} d x

Apply u-substitution

: \int \frac{e ^{- \frac{6}{u}}}{4 u ^{2}} d u

= 2 \cdot \int \frac{e ^{- \frac{6}{u}}}{4 u ^{2}} d u

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int \frac{e ^{- \frac{6}{u}}}{u ^{2}} d u

Apply u-substitution

: \int \frac{e ^{v}}{6} d v

= 2 \cdot \frac{1}{4} \cdot \int \frac{e ^{v}}{6} d v

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \frac{1}{6} \cdot \int e^{v} d v

Use the common integral:

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

= 2 \cdot \frac{1}{4} \cdot \frac{1}{6} e^{v}

Substitute back

= 2 \cdot \frac{1}{4} \cdot \frac{1}{6} e^{- \frac{6}{x ^{4}}}

Simplify

2 \cdot \frac{1}{4} \cdot \frac{1}{6} e^{- \frac{6}{x ^{4}}} : \frac{1}{12} e^{- \frac{6}{x ^{4}}}

= \frac{1}{12} e^{- \frac{6}{x ^{4}}}

Add a constant to the solution

= \frac{1}{12} e^{- \frac{6}{x ^{4}}} + C

\frac{d ^{2} y}{d t ^{2}} - 8 \frac{d y}{d t} + 16 y = 0

f (x) = x^{3}, at x = 1

y = ln (18 x^{2} - 5 x + 11)

2^{x + 3}

\int \frac{x ^{3}}{3 x ^{2} + 3} d x

What is the integral of (2e^{-6/(x^4)})/(x^5) ?
The integral of (2e^{-6/(x^4)})/(x^5) is 1/12 e^{-6/(x^4)}+C