What is the integral of (8x^3)/(e^{x^4)} ?

The integral of (8x^3)/(e^{x^4)} is -2e^{-x^4}+C

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integral of (8x^3)/(e^{x^4)}

\int \frac{8 x ^{3}}{e ^{x^{4}}} d x

- 2 e^{- x^{4}} + C

Solution steps

\int \frac{8 x ^{3}}{e ^{x^{4}}} d x

Take the constant out:

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

= 8 \cdot \int \frac{x ^{3}}{e ^{x^{4}}} d x

Apply exponent rule:

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

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

= 8 \cdot \int x^{3} e^{- x^{4}} d x

Apply u-substitution

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

= 8 \cdot \int - \frac{e ^{u}}{4} d u

Take the constant out:

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

= 8 (- \frac{1}{4} \cdot \int e^{u} d u)

Use the common integral:

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

= 8 (- \frac{1}{4} e^{u})

Substitute back

u = - x^{4}

= 8 (- \frac{1}{4} e^{- x^{4}})

Simplify

8 (- \frac{1}{4} e^{- x^{4}}) : - 2 e^{- x^{4}}

= - 2 e^{- x^{4}}

Add a constant to the solution

= - 2 e^{- x^{4}} + C

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

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\int_{4}^{14} 16 + (4 - x)^{2} d x

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\int_{0}^{1.5} 70 x d x

What is the integral of (8x^3)/(e^{x^4)} ?
The integral of (8x^3)/(e^{x^4)} is -2e^{-x^4}+C