What is the integral of ye^{3y^2} ?

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

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integral of ye^{3y^2}

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

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

Solution steps

\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

\frac{d}{d x} (I n x^{2})

\frac{d}{d x} (ln (x^{2} - 3 x + 2))

\frac{d}{d x} (\frac{x}{4 + x})

3 y^{^{''}} + 8 y^{^{'}} - 16 y = 0

y^{^{''}} + 4 y = e^{- x}