What is the integral of (x^5)/(e^{2-3x^6)} ?

The integral of (x^5)/(e^{2-3x^6)} is 1/18 e^{3x^6-2}+C

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integral of (x^5)/(e^{2-3x^6)}

\int \frac{x ^{5}}{e ^{2 - 3 x^{6}}} d x

\frac{1}{18} e^{3 x^{6} - 2} + C

Solution steps

\int \frac{x ^{5}}{e ^{2 - 3 x^{6}}} d x

\frac{x ^{5}}{e ^{2 - 3 x^{6}}} = \frac{1}{e ^{2}} \frac{x ^{5}}{e ^{- 3 x^{6}}}

= \int \frac{1}{e ^{2}} \cdot \frac{x ^{5}}{e ^{- 3 x^{6}}} d x

Take the constant out:

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

= \frac{1}{e ^{2}} \cdot \int \frac{x ^{5}}{e ^{- 3 x^{6}}} d x

Simplify

\frac{x ^{5}}{e ^{- 3 x^{6}}} : e^{3 x^{6}} x^{5}

= \frac{1}{e ^{2}} \cdot \int e^{3 x^{6}} x^{5} d x

Apply u-substitution

: \int \frac{e ^{3 u}}{6} d u

= \frac{1}{e ^{2}} \cdot \int \frac{e ^{3 u}}{6} d u

Take the constant out:

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

= \frac{1}{e ^{2}} \cdot \frac{1}{6} \cdot \int e^{3 u} d u

Apply u-substitution

: \int e^{v} \frac{1}{3} d v

= \frac{1}{e ^{2}} \cdot \frac{1}{6} \cdot \int e^{v} \frac{1}{3} d v

Take the constant out:

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

= \frac{1}{e ^{2}} \cdot \frac{1}{6} \cdot \frac{1}{3} \cdot \int e^{v} d v

Use the common integral:

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

= \frac{1}{e ^{2}} \cdot \frac{1}{6} \cdot \frac{1}{3} e^{v}

Substitute back

= \frac{1}{e ^{2}} \cdot \frac{1}{6} \cdot \frac{1}{3} e^{3 x^{6}}

Simplify

\frac{1}{e ^{2}} \cdot \frac{1}{6} \cdot \frac{1}{3} e^{3 x^{6}} : \frac{1}{18} e^{3 x^{6} - 2}

= \frac{1}{18} e^{3 x^{6} - 2}

Add a constant to the solution

= \frac{1}{18} e^{3 x^{6} - 2} + C

n = 1 \sum \infty \frac{1}{1 - 2 n}

3 t

\int x^{2} (x - 2)^{\frac{1}{2}} d x

\frac{d}{d x} (5 cos (x))

\frac{d}{d x} ((x + 1)^{x^{2} + x^{3}})

What is the integral of (x^5)/(e^{2-3x^6)} ?
The integral of (x^5)/(e^{2-3x^6)} is 1/18 e^{3x^6-2}+C