Resolva as integrais a seguir usando o método de integração por partes.

[tex]\displaystyle \int \text{x.e}^{\text x}\text{dx}= \text {u.v}-\int \text v\text{du} \\\\ \underline{\text{fa{\c c}amos}}: \\\\ \text{dv}= \text e^{\text x}\text{dx} \to \text v = \text e^{\text x} \\\\ \text u = \text x \to \text{du} = 1\text{dx}\\\\ \underline{\text{substituindo}}: \\\\ \int \text x.\text e^{\text x}\text {dx} = \text x.\text e^{\text x} - \int \text e^{\text x}.\text{dx} \\\\\\ \huge\boxed{\int \text{x.e}^{\text x}\text{dx} = \text x.\text{e}^{\text x}-\text e^{\text x}+\text C\ }[/tex]

item b)

[tex]\displaystyle \int \text x.\text{ln x}\ {\text{dx}} = \int \text{u.dv} = \text{u.v}-\int \text{v.du} \\\\ \underline{\text{fa{\c c}amos}}: \\\\ \text {dv = x } \to \text{v}=\frac{\text x^2}{2}\text{ \dx}\\\\ \text{u = ln x} \to \text{du }=\frac{1}{\text x}\text{dx} \\\\\underline{\text{substituindo}}: \\\\ \int\text{x.ln x dx} = \frac{\text x^2.\text{ln x}}{2} - \int \frac{\text x^2}{2}.\frac{1}{\text x}\text{dx}[/tex]

[tex]\displaystyle \int\text{x.ln x dx} = \frac{\text x^2.\text{ln x}}{2} - \int \frac{\text x}{2}\text{dx} \\\\\\ \huge\boxed{\int\text{x.ln x dx} = \frac{\text x^2.\text{ln x}}{2} - \frac{\text x^2}{4}+\text C}\checkmark[/tex]

item c)

[tex]\displaystyle \int \text x.\text{ln x}\ {\text{dx}} = \int \text{u.dv} = \text{u.v}-\int \text{v.du} \\\\ \underline{\text{fa{\c c}amos}}: \\\\ \text {dv = x }^2 \to \text{v}=\frac{\text x^3}{3}\text{ \dx}\\\\ \text{u = ln x} \to \text{du }=\frac{1}{\text x}\text{dx} \\\\\underline{\text{substituindo}}: \\\\ \int\text{x}^2\text{ln x dx} = \frac{\text x^3.\text{ln x}}{3} - \int \frac{\text x^3}{3}.\frac{1}{\text x}\text{dx}[/tex]

[tex]\displaystyle \int\text{x.ln x dx} = \frac{\text x^3.\text{ln x}}{3} - \int \frac{\text x^2}{3}\text{dx} \\\\\\ \huge\boxed{\int\text{x.ln x dx} = \frac{\text x^3.\text{ln x}}{3} - \frac{\text x^3}{9}+\text C}\checkmark[/tex]