Question

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

Answer 1

a)

∫ x * e^(x)  dx

u=x  ==> du =dx

dv=e^(x)  dx   ==> ∫dv=∫e^(x)  dx  ==>v=e^(x)

∫ x * e^(x)  dx =x* e^(x)- ∫e^(x) dx

∫ x * e^(x)  dx =x* e^(x)- e^(x)  + c

b)

∫ x * ln(x)  dx

u= ln(x)  ==>du= dx/x

dv =x dx ==>∫ dv =  ∫ x  dv  ==> v= x²/2

∫ x * ln(x)  dx= (x²/2) * ln(x)- ∫ x²/2   dx/x

∫ x * ln(x)  dx= (x²/2) * ln(x)- (1/2)*∫ x   dx

∫ x * ln(x)  dx= (x²/2) * ln(x)- (1/2)*x²/2 + c

∫ x * ln(x)  dx= (x²/2) * ln(x)- (1/4)*x²+ c

c)

∫ x² * ln(x)  dx

u=ln(x)  ==> du=dx/x

dv=x²   dx    ==>∫ dv=∫x²   dx ==>v= x³/3

∫ x² * ln(x)  dx = ln(x) * x³/3 - ∫ x³/3  * dx/x

∫ x² * ln(x)  dx = ln(x) * x³/3 -(1/3)* ∫ x²  * dx

∫ x² * ln(x)  dx = ln(x) * x³/3 -(1/3)* x³/3  + c

∫ x² * ln(x)  dx = ln(x) * x³/3 -(1/9)* x³  + c

Answer 2

integração por partes :

$\displaystyle \int \text u.\text{dv} = \text {u.v} - \int \text v\ \text{du}$

item a)

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

item b)

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

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

item c)

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

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