What is the integral of (4x^2)/(e^{x^3)} ?

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

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integral of (4x^2)/(e^{x^3)}

\int \frac{4 x ^{2}}{e ^{x^{3}}} d x

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

Solution steps

\int \frac{4 x ^{2}}{e ^{x^{3}}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{2}}{e ^{x^{3}}} d x

Apply exponent rule:

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

\frac{x ^{2}}{e ^{x^{3}}} = (x^{2}) e^{- x^{3}}

= 4 \cdot \int x^{2} e^{- x^{3}} d x

Apply u-substitution

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

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

Take the constant out:

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

= 4 (- \frac{1}{3} \cdot \int e^{u} d u)

Use the common integral:

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

= 4 (- \frac{1}{3} e^{u})

Substitute back

u = - x^{3}

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

Simplify

4 (- \frac{1}{3} e^{- x^{3}}) : - \frac{4}{3} e^{- x^{3}}

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

Add a constant to the solution

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

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