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

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

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integral of e^{y^2}y^3

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

\frac{1}{2} (e^{y^{2}} y^{2} - e^{y^{2}}) + C

Solution steps

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

Apply u-substitution

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

= \int \frac{e ^{u} u}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{2} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{2} (e^{u} u - e^{u})

Substitute back

u = y^{2}

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

Add a constant to the solution

= \frac{1}{2} (e^{y^{2}} y^{2} - e^{y^{2}}) + C

(7 x + 6)^{2}

\int 8 x^{\frac{1}{3}} - 4 x^{- \frac{1}{3}} d x

x \to - \infty lim (\frac{2 e ^{- \frac{1}{2} x}}{x ^{3}})

\int 2 cos^{5} (x) sin^{4} (x) d x

f (x) = - \frac{9 x ^{2}}{2} + 8 x - 22

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