Question

Resolva a integral: integral (x^3 + 2/raiz de X )^2 dx

Answer 1

Olá

$\displaystyle \int ( \frac{x^3+2}{ \sqrt{x} } )^2dx \\ \\ \\ u=x^3 \\ du=3x^2dx \\ \\dx= \frac{du}{3x^2} ~~~~ ~~ \longrightarrow~ ~~dx= \frac{1}{3x^2} du \\ \\ \text{substituindo na integral} \\ \\ \displaystyle \int ( \frac{u+2}{ \sqrt{x} } )^2~ \frac{1}{3x^2} du \\ \\ \\ \text{cancela a raiz quadrada do denominador com o expoente} \\ \\ \\ \displaystyle \int \frac{(u+2)^2}{ x } \cdot\frac{1}{3x^2}du \\ \\ \\ \displaystyle \int \frac{(u+2)^2}{ x\cdot3x^2 } du$

$\displaystyle \int \frac{(u+2)^2}{ 3x^3 } du \\ \\ \\ \frac{1}{3} \displaystyle \int \frac{(u+2)^2}{ x^3 } du \\ \\ \\ \text{temos que}~x^3~\text{ e "u", entao substituindo} \\ \\ \\ \frac{1}{3} \displaystyle \int \frac{(u+2)^2}{ u } du \\ \\ \\ \text{expande o termo}~(u+2)^2 \\ \\ (u+2)^2=u^2+4u+u \\ \\ \\ \frac{1}{3} \displaystyle \int \frac{u^2+4u+4}{ u } du$

$\text{Separando e fazendo as devidas simplificacoes} \\ \\ \frac{1}{3}( \displaystyle \int \frac{u^\diagup\!\!\!\!^2}{ \diagup\!\!\!\!\!u } du~~+~~\int \frac{4\diagup\!\!\!\!u}{\diagup\!\!\!\!u}du ~+~\int \frac{4}{u}du ) \\ \\ \\ \frac{1}{3} (\displaystyle \int udu~~+~~\int 4du~~+~~\int \frac{4}{u} du) \\ \\ \\ \frac{1}{3}~( \frac{u^2}{2} ~+~ 4u~+~4ln|u|} ) \\ \\ \\ \text{substituindo o "u" de volta por}~x^3 \\ \\ \\ \frac{1}{3}~( \frac{(x^3)^2}{2} ~+~ 4x^3~+~4ln|x^3|} )$

$\text{Pelas propriedade do logaritmo natural} \\ \\ ln(a^b)=b\cdot ln(a)$

$\displaystyle \frac{1}{3}~( \frac{x^6}{2} ~+~ 4x^3~+~3\cdot4ln|x|} ) \\ \\ \\\boxed{\boxed{ \frac{1}{3}~( \frac{x^6}{2} ~+~ 4x^3~+~12ln|x| )+C}}$






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