Integraal van $$$x^{2} \sqrt{e^{x^{3}} + 5} e^{x^{3}}$$$

Bepaal $$$\int x^{2} \sqrt{e^{x^{3}} + 5} e^{x^{3}}\, dx$$$.

Zij $$$u=x^{3}$$$.

Dan $$$du=\left(x^{3}\right)^{\prime }dx = 3 x^{2} dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$x^{2} dx = \frac{du}{3}$$$.

Dus,

$${\color{red}{\int{x^{2} \sqrt{e^{x^{3}} + 5} e^{x^{3}} d x}}} = {\color{red}{\int{\frac{\sqrt{e^{u} + 5} e^{u}}{3} d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{3}$$$ en $$$f{\left(u \right)} = \sqrt{e^{u} + 5} e^{u}$$$:

$${\color{red}{\int{\frac{\sqrt{e^{u} + 5} e^{u}}{3} d u}}} = {\color{red}{\left(\frac{\int{\sqrt{e^{u} + 5} e^{u} d u}}{3}\right)}}$$

Zij $$$v=e^{u} + 5$$$.

Dan $$$dv=\left(e^{u} + 5\right)^{\prime }du = e^{u} du$$$ (de stappen zijn te zien »), en dan geldt dat $$$e^{u} du = dv$$$.

Dus,

$$\frac{{\color{red}{\int{\sqrt{e^{u} + 5} e^{u} d u}}}}{3} = \frac{{\color{red}{\int{\sqrt{v} d v}}}}{3}$$

Pas de machtsregel $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=\frac{1}{2}$$$:

$$\frac{{\color{red}{\int{\sqrt{v} d v}}}}{3}=\frac{{\color{red}{\int{v^{\frac{1}{2}} d v}}}}{3}=\frac{{\color{red}{\frac{v^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}}{3}=\frac{{\color{red}{\left(\frac{2 v^{\frac{3}{2}}}{3}\right)}}}{3}$$

We herinneren eraan dat $$$v=e^{u} + 5$$$:

$$\frac{2 {\color{red}{v}}^{\frac{3}{2}}}{9} = \frac{2 {\color{red}{\left(e^{u} + 5\right)}}^{\frac{3}{2}}}{9}$$

We herinneren eraan dat $$$u=x^{3}$$$:

$$\frac{2 \left(5 + e^{{\color{red}{u}}}\right)^{\frac{3}{2}}}{9} = \frac{2 \left(5 + e^{{\color{red}{x^{3}}}}\right)^{\frac{3}{2}}}{9}$$

Dus,

$$\int{x^{2} \sqrt{e^{x^{3}} + 5} e^{x^{3}} d x} = \frac{2 \left(e^{x^{3}} + 5\right)^{\frac{3}{2}}}{9}$$

Voeg de integratieconstante toe:

$$\int{x^{2} \sqrt{e^{x^{3}} + 5} e^{x^{3}} d x} = \frac{2 \left(e^{x^{3}} + 5\right)^{\frac{3}{2}}}{9}+C$$

$$$\int x^{2} \sqrt{e^{x^{3}} + 5} e^{x^{3}}\, dx = \frac{2 \left(e^{x^{3}} + 5\right)^{\frac{3}{2}}}{9} + C$$$A