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

The integral of (x^3)/(e^{3x^4)} is -1/12 e^{-3x^4}+C

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

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

- \frac{1}{12} e^{- 3 x^{4}} + C

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{x ^{3}}{e ^{3 x^{4}}} = (x^{3}) e^{- 3 x^{4}}

= \int x^{3} e^{- 3 x^{4}} d x

Apply u-substitution

: \int - \frac{e ^{u}}{12} d u

= \int - \frac{e ^{u}}{12} d u

Take the constant out:

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

= - \frac{1}{12} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{12} e^{u}

Substitute back

u = - 3 x^{4}

= - \frac{1}{12} e^{- 3 x^{4}}

Add a constant to the solution

= - \frac{1}{12} e^{- 3 x^{4}} + C

e^{x}, at x = 3

\int_{0}^{1} x^{3} 1 - x^{2} d x

- \frac{2}{3 x ^{2}}

\frac{\partial}{\partial x} (2 x y - 3 y^{2})

\frac{d}{d x} ((x^{3} + 3) 2^{- 7 x})

What is the integral of (x^3)/(e^{3x^4)} ?
The integral of (x^3)/(e^{3x^4)} is -1/12 e^{-3x^4}+C