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

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

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integral of (5y^2)/(e^{2y^3)}

\int \frac{5 y ^{2}}{e ^{2 y^{3}}} d y

- \frac{5}{6} e^{- 2 y^{3}} + C

Solution steps

\int \frac{5 y ^{2}}{e ^{2 y^{3}}} d y

Take the constant out:

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

= 5 \cdot \int \frac{y ^{2}}{e ^{2 y^{3}}} d y

Apply exponent rule:

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

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

= 5 \cdot \int y^{2} e^{- 2 y^{3}} d y

Apply u-substitution

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

= 5 \cdot \int - \frac{e ^{u}}{6} d u

Take the constant out:

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

= 5 (- \frac{1}{6} \cdot \int e^{u} d u)

Use the common integral:

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

= 5 (- \frac{1}{6} e^{u})

Substitute back

u = - 2 y^{3}

= 5 (- \frac{1}{6} e^{- 2 y^{3}})

Simplify

5 (- \frac{1}{6} e^{- 2 y^{3}}) : - \frac{5}{6} e^{- 2 y^{3}}

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

Add a constant to the solution

= - \frac{5}{6} e^{- 2 y^{3}} + C

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