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

The integral of (3y^2)/(4e^{-y^3)} is 1/4 e^{y^3}+C

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

\int \frac{3 y ^{2}}{4 e ^{- y^{3}}} d y

\frac{1}{4} e^{y^{3}} + C

Solution steps

\int \frac{3 y ^{2}}{4 e ^{- y^{3}}} d y

Take the constant out:

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

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

Apply u-substitution

: \int \frac{e ^{u^{\frac{3}{2}}} u}{2 y} d u

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

Take the constant out:

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

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

Simplify

\frac{1}{e ^{- u}} : e^{u}

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

Use the common integral:

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

= \frac{3}{4} \cdot \frac{1}{3} e^{u}

Substitute back

u = y^{3}

= \frac{3}{4} \cdot \frac{1}{3} e^{y^{3}}

Simplify

\frac{3}{4} \cdot \frac{1}{3} e^{y^{3}} : \frac{1}{4} e^{y^{3}}

= \frac{1}{4} e^{y^{3}}

Add a constant to the solution

= \frac{1}{4} e^{y^{3}} + C

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What is the integral of (3y^2)/(4e^{-y^3)} ?
The integral of (3y^2)/(4e^{-y^3)} is 1/4 e^{y^3}+C