Integrale di $$$x^{2} \operatorname{atan}{\left(\sqrt{x} \right)}$$$

Trova $$$\int x^{2} \operatorname{atan}{\left(\sqrt{x} \right)}\, dx$$$.

Per l'integrale $$$\int{x^{2} \operatorname{atan}{\left(\sqrt{x} \right)} d x}$$$, usa l'integrazione per parti $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Siano $$$\operatorname{u}=\operatorname{atan}{\left(\sqrt{x} \right)}$$$ e $$$\operatorname{dv}=x^{2} dx$$$.

Quindi $$$\operatorname{du}=\left(\operatorname{atan}{\left(\sqrt{x} \right)}\right)^{\prime }dx=\frac{1}{2 \sqrt{x} \left(x + 1\right)} dx$$$ (i passaggi si possono vedere ») e $$$\operatorname{v}=\int{x^{2} d x}=\frac{x^{3}}{3}$$$ (i passaggi si possono vedere »).

Pertanto,

$${\color{red}{\int{x^{2} \operatorname{atan}{\left(\sqrt{x} \right)} d x}}}={\color{red}{\left(\operatorname{atan}{\left(\sqrt{x} \right)} \cdot \frac{x^{3}}{3}-\int{\frac{x^{3}}{3} \cdot \frac{1}{2 \sqrt{x} \left(x + 1\right)} d x}\right)}}={\color{red}{\left(\frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} - \int{\frac{x^{\frac{5}{2}}}{6 x + 6} d x}\right)}}$$

Semplifica l’integranda:

$$\frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} - {\color{red}{\int{\frac{x^{\frac{5}{2}}}{6 x + 6} d x}}} = \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} - {\color{red}{\int{\frac{x^{\frac{5}{2}}}{6 \left(x + 1\right)} d x}}}$$

Applica la regola del fattore costante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{1}{6}$$$ e $$$f{\left(x \right)} = \frac{x^{\frac{5}{2}}}{x + 1}$$$:

$$\frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} - {\color{red}{\int{\frac{x^{\frac{5}{2}}}{6 \left(x + 1\right)} d x}}} = \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} - {\color{red}{\left(\frac{\int{\frac{x^{\frac{5}{2}}}{x + 1} d x}}{6}\right)}}$$

Sia $$$u=\sqrt{x}$$$.

Quindi $$$du=\left(\sqrt{x}\right)^{\prime }dx = \frac{1}{2 \sqrt{x}} dx$$$ (i passaggi si possono vedere »), e si ha che $$$\frac{dx}{\sqrt{x}} = 2 du$$$.

L'integrale può essere riscritto come

$$\frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} - \frac{{\color{red}{\int{\frac{x^{\frac{5}{2}}}{x + 1} d x}}}}{6} = \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} - \frac{{\color{red}{\int{\frac{2 u^{6}}{u^{2} + 1} d u}}}}{6}$$

Applica la regola del fattore costante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=2$$$ e $$$f{\left(u \right)} = \frac{u^{6}}{u^{2} + 1}$$$:

$$\frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} - \frac{{\color{red}{\int{\frac{2 u^{6}}{u^{2} + 1} d u}}}}{6} = \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} - \frac{{\color{red}{\left(2 \int{\frac{u^{6}}{u^{2} + 1} d u}\right)}}}{6}$$

Poiché il grado del numeratore non è inferiore al grado del denominatore, esegui la divisione lunga dei polinomi (i passaggi sono visibili »):

$$\frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} - \frac{{\color{red}{\int{\frac{u^{6}}{u^{2} + 1} d u}}}}{3} = \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} - \frac{{\color{red}{\int{\left(u^{4} - u^{2} + 1 - \frac{1}{u^{2} + 1}\right)d u}}}}{3}$$

Integra termine per termine:

$$\frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} - \frac{{\color{red}{\int{\left(u^{4} - u^{2} + 1 - \frac{1}{u^{2} + 1}\right)d u}}}}{3} = \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} - \frac{{\color{red}{\left(\int{1 d u} - \int{u^{2} d u} + \int{u^{4} d u} - \int{\frac{1}{u^{2} + 1} d u}\right)}}}{3}$$

Applica la regola della costante $$$\int c\, du = c u$$$ con $$$c=1$$$:

$$\frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} + \frac{\int{u^{2} d u}}{3} - \frac{\int{u^{4} d u}}{3} + \frac{\int{\frac{1}{u^{2} + 1} d u}}{3} - \frac{{\color{red}{\int{1 d u}}}}{3} = \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} + \frac{\int{u^{2} d u}}{3} - \frac{\int{u^{4} d u}}{3} + \frac{\int{\frac{1}{u^{2} + 1} d u}}{3} - \frac{{\color{red}{u}}}{3}$$

Applica la regola della potenza $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=4$$$:

$$- \frac{u}{3} + \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} + \frac{\int{u^{2} d u}}{3} + \frac{\int{\frac{1}{u^{2} + 1} d u}}{3} - \frac{{\color{red}{\int{u^{4} d u}}}}{3}=- \frac{u}{3} + \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} + \frac{\int{u^{2} d u}}{3} + \frac{\int{\frac{1}{u^{2} + 1} d u}}{3} - \frac{{\color{red}{\frac{u^{1 + 4}}{1 + 4}}}}{3}=- \frac{u}{3} + \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} + \frac{\int{u^{2} d u}}{3} + \frac{\int{\frac{1}{u^{2} + 1} d u}}{3} - \frac{{\color{red}{\left(\frac{u^{5}}{5}\right)}}}{3}$$

Applica la regola della potenza $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=2$$$:

$$- \frac{u^{5}}{15} - \frac{u}{3} + \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} + \frac{\int{\frac{1}{u^{2} + 1} d u}}{3} + \frac{{\color{red}{\int{u^{2} d u}}}}{3}=- \frac{u^{5}}{15} - \frac{u}{3} + \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} + \frac{\int{\frac{1}{u^{2} + 1} d u}}{3} + \frac{{\color{red}{\frac{u^{1 + 2}}{1 + 2}}}}{3}=- \frac{u^{5}}{15} - \frac{u}{3} + \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} + \frac{\int{\frac{1}{u^{2} + 1} d u}}{3} + \frac{{\color{red}{\left(\frac{u^{3}}{3}\right)}}}{3}$$

L'integrale di $$$\frac{1}{u^{2} + 1}$$$ è $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

$$- \frac{u^{5}}{15} + \frac{u^{3}}{9} - \frac{u}{3} + \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} + \frac{{\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}}{3} = - \frac{u^{5}}{15} + \frac{u^{3}}{9} - \frac{u}{3} + \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} + \frac{{\color{red}{\operatorname{atan}{\left(u \right)}}}}{3}$$

Ricordiamo che $$$u=\sqrt{x}$$$:

$$\frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} + \frac{\operatorname{atan}{\left({\color{red}{u}} \right)}}{3} - \frac{{\color{red}{u}}}{3} + \frac{{\color{red}{u}}^{3}}{9} - \frac{{\color{red}{u}}^{5}}{15} = \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} + \frac{\operatorname{atan}{\left({\color{red}{\sqrt{x}}} \right)}}{3} - \frac{{\color{red}{\sqrt{x}}}}{3} + \frac{{\color{red}{\sqrt{x}}}^{3}}{9} - \frac{{\color{red}{\sqrt{x}}}^{5}}{15}$$

Pertanto,

$$\int{x^{2} \operatorname{atan}{\left(\sqrt{x} \right)} d x} = - \frac{x^{\frac{5}{2}}}{15} + \frac{x^{\frac{3}{2}}}{9} - \frac{\sqrt{x}}{3} + \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} + \frac{\operatorname{atan}{\left(\sqrt{x} \right)}}{3}$$

Aggiungi la costante di integrazione:

$$\int{x^{2} \operatorname{atan}{\left(\sqrt{x} \right)} d x} = - \frac{x^{\frac{5}{2}}}{15} + \frac{x^{\frac{3}{2}}}{9} - \frac{\sqrt{x}}{3} + \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} + \frac{\operatorname{atan}{\left(\sqrt{x} \right)}}{3}+C$$

$$$\int x^{2} \operatorname{atan}{\left(\sqrt{x} \right)}\, dx = \left(- \frac{x^{\frac{5}{2}}}{15} + \frac{x^{\frac{3}{2}}}{9} - \frac{\sqrt{x}}{3} + \frac{x^{3} \operatorname{atan}{\left(\sqrt{x} \right)}}{3} + \frac{\operatorname{atan}{\left(\sqrt{x} \right)}}{3}\right) + C$$$A