Integral of $$$e^{\sqrt[3]{x}}$$$

Find $$$\int e^{\sqrt[3]{x}}\, dx$$$.

Let $$$u=\sqrt[3]{x}$$$.

Then $$$du=\left(\sqrt[3]{x}\right)^{\prime }dx = \frac{1}{3 x^{\frac{2}{3}}} dx$$$ (steps can be seen »), and we have that $$$\frac{dx}{x^{\frac{2}{3}}} = 3 du$$$.

The integral becomes

$${\color{red}{\int{e^{\sqrt[3]{x}} d x}}} = {\color{red}{\int{3 u^{2} e^{u} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=3$$$ and $$$f{\left(u \right)} = u^{2} e^{u}$$$:

$${\color{red}{\int{3 u^{2} e^{u} d u}}} = {\color{red}{\left(3 \int{u^{2} e^{u} d u}\right)}}$$

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

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

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

The integral can be rewritten as

$$3 {\color{red}{\int{u^{2} e^{u} d u}}}=3 {\color{red}{\left(u^{2} \cdot e^{u}-\int{e^{u} \cdot 2 u d u}\right)}}=3 {\color{red}{\left(u^{2} e^{u} - \int{2 u e^{u} d u}\right)}}$$

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

$$3 u^{2} e^{u} - 3 {\color{red}{\int{2 u e^{u} d u}}} = 3 u^{2} e^{u} - 3 {\color{red}{\left(2 \int{u e^{u} d u}\right)}}$$

For the integral $$$\int{u e^{u} d u}$$$, use integration by parts $$$\int \operatorname{c} \operatorname{dv} = \operatorname{c}\operatorname{v} - \int \operatorname{v} \operatorname{dc}$$$.

Let $$$\operatorname{c}=u$$$ and $$$\operatorname{dv}=e^{u} du$$$.

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

So,

$$3 u^{2} e^{u} - 6 {\color{red}{\int{u e^{u} d u}}}=3 u^{2} e^{u} - 6 {\color{red}{\left(u \cdot e^{u}-\int{e^{u} \cdot 1 d u}\right)}}=3 u^{2} e^{u} - 6 {\color{red}{\left(u e^{u} - \int{e^{u} d u}\right)}}$$

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

$$3 u^{2} e^{u} - 6 u e^{u} + 6 {\color{red}{\int{e^{u} d u}}} = 3 u^{2} e^{u} - 6 u e^{u} + 6 {\color{red}{e^{u}}}$$

Recall that $$$u=\sqrt[3]{x}$$$:

$$6 e^{{\color{red}{u}}} - 6 {\color{red}{u}} e^{{\color{red}{u}}} + 3 {\color{red}{u}}^{2} e^{{\color{red}{u}}} = 6 e^{{\color{red}{\sqrt[3]{x}}}} - 6 {\color{red}{\sqrt[3]{x}}} e^{{\color{red}{\sqrt[3]{x}}}} + 3 {\color{red}{\sqrt[3]{x}}}^{2} e^{{\color{red}{\sqrt[3]{x}}}}$$

Therefore,

$$\int{e^{\sqrt[3]{x}} d x} = 3 x^{\frac{2}{3}} e^{\sqrt[3]{x}} - 6 \sqrt[3]{x} e^{\sqrt[3]{x}} + 6 e^{\sqrt[3]{x}}$$

Simplify:

$$\int{e^{\sqrt[3]{x}} d x} = 3 \left(x^{\frac{2}{3}} - 2 \sqrt[3]{x} + 2\right) e^{\sqrt[3]{x}}$$

Add the constant of integration:

$$\int{e^{\sqrt[3]{x}} d x} = 3 \left(x^{\frac{2}{3}} - 2 \sqrt[3]{x} + 2\right) e^{\sqrt[3]{x}}+C$$

$$$\int e^{\sqrt[3]{x}}\, dx = 3 \left(x^{\frac{2}{3}} - 2 \sqrt[3]{x} + 2\right) e^{\sqrt[3]{x}} + C$$$A