Integral of $$$y^{3} e^{\frac{y^{2}}{2}}$$$

Find $$$\int y^{3} e^{\frac{y^{2}}{2}}\, dy$$$.

Let $$$u=y^{2}$$$.

Then $$$du=\left(y^{2}\right)^{\prime }dy = 2 y dy$$$ (steps can be seen »), and we have that $$$y dy = \frac{du}{2}$$$.

So,

$${\color{red}{\int{y^{3} e^{\frac{y^{2}}{2}} d y}}} = {\color{red}{\int{\frac{u e^{\frac{u}{2}}}{2} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = u e^{\frac{u}{2}}$$$:

$${\color{red}{\int{\frac{u e^{\frac{u}{2}}}{2} d u}}} = {\color{red}{\left(\frac{\int{u e^{\frac{u}{2}} d u}}{2}\right)}}$$

For the integral $$$\int{u e^{\frac{u}{2}} d u}$$$, use integration by parts $$$\int \operatorname{m} \operatorname{dv} = \operatorname{m}\operatorname{v} - \int \operatorname{v} \operatorname{dm}$$$.

Let $$$\operatorname{m}=u$$$ and $$$\operatorname{dv}=e^{\frac{u}{2}} du$$$.

Then $$$\operatorname{dm}=\left(u\right)^{\prime }du=1 du$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{\frac{u}{2}} d u}=2 e^{\frac{u}{2}}$$$ (steps can be seen »).

The integral can be rewritten as

$$\frac{{\color{red}{\int{u e^{\frac{u}{2}} d u}}}}{2}=\frac{{\color{red}{\left(u \cdot 2 e^{\frac{u}{2}}-\int{2 e^{\frac{u}{2}} \cdot 1 d u}\right)}}}{2}=\frac{{\color{red}{\left(2 u e^{\frac{u}{2}} - \int{2 e^{\frac{u}{2}} d u}\right)}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=2$$$ and $$$f{\left(u \right)} = e^{\frac{u}{2}}$$$:

$$u e^{\frac{u}{2}} - \frac{{\color{red}{\int{2 e^{\frac{u}{2}} d u}}}}{2} = u e^{\frac{u}{2}} - \frac{{\color{red}{\left(2 \int{e^{\frac{u}{2}} d u}\right)}}}{2}$$

Let $$$v=\frac{u}{2}$$$.

Then $$$dv=\left(\frac{u}{2}\right)^{\prime }du = \frac{du}{2}$$$ (steps can be seen »), and we have that $$$du = 2 dv$$$.

Therefore,

$$u e^{\frac{u}{2}} - {\color{red}{\int{e^{\frac{u}{2}} d u}}} = u e^{\frac{u}{2}} - {\color{red}{\int{2 e^{v} d v}}}$$

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=2$$$ and $$$f{\left(v \right)} = e^{v}$$$:

$$u e^{\frac{u}{2}} - {\color{red}{\int{2 e^{v} d v}}} = u e^{\frac{u}{2}} - {\color{red}{\left(2 \int{e^{v} d v}\right)}}$$

The integral of the exponential function is $$$\int{e^{v} d v} = e^{v}$$$:

$$u e^{\frac{u}{2}} - 2 {\color{red}{\int{e^{v} d v}}} = u e^{\frac{u}{2}} - 2 {\color{red}{e^{v}}}$$

Recall that $$$v=\frac{u}{2}$$$:

$$u e^{\frac{u}{2}} - 2 e^{{\color{red}{v}}} = u e^{\frac{u}{2}} - 2 e^{{\color{red}{\left(\frac{u}{2}\right)}}}$$

Recall that $$$u=y^{2}$$$:

$$- 2 e^{\frac{{\color{red}{u}}}{2}} + {\color{red}{u}} e^{\frac{{\color{red}{u}}}{2}} = - 2 e^{\frac{{\color{red}{y^{2}}}}{2}} + {\color{red}{y^{2}}} e^{\frac{{\color{red}{y^{2}}}}{2}}$$

Therefore,

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

Simplify:

$$\int{y^{3} e^{\frac{y^{2}}{2}} d y} = \left(y^{2} - 2\right) e^{\frac{y^{2}}{2}}$$

Add the constant of integration:

$$\int{y^{3} e^{\frac{y^{2}}{2}} d y} = \left(y^{2} - 2\right) e^{\frac{y^{2}}{2}}+C$$

$$$\int y^{3} e^{\frac{y^{2}}{2}}\, dy = \left(y^{2} - 2\right) e^{\frac{y^{2}}{2}} + C$$$A