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

The integral of y/(e^{3y^2)} is -1/6 e^{-3y^2}+C

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

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

- \frac{1}{6} e^{- 3 y^{2}} + C

Solution steps

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

Apply exponent rule:

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

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

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

Apply u-substitution

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

= \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

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

Use the common integral:

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

= - \frac{1}{6} e^{u}

Substitute back

u = - 3 y^{2}

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

Add a constant to the solution

= - \frac{1}{6} e^{- 3 y^{2}} + C

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What is the integral of y/(e^{3y^2)} ?
The integral of y/(e^{3y^2)} is -1/6 e^{-3y^2}+C