Question

Integral de x^3*(x^2+1)^(3/2)

Answer 1

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Calcular a integral indefinida:

$\mathsf{\displaystyle\int\! x^3\cdot (x^2+1)^{3/2}\,dx}$


Vamos reescrever a função integrando de forma conveniente:

$\mathsf{=\displaystyle\int\! x\cdot x^2\cdot (x^2+1)^{3/2}\,dx}\\\\\\ \mathsf{=\displaystyle\int\! \frac{1}{2}\cdot 2x\cdot x^2\cdot (x^2+1)^{3/2}\,dx}\\\\\\ \mathsf{=\displaystyle\frac{1}{2}\int\! x^2\cdot (x^2+1)^{3/2}\cdot 2x\,dx}$


Agora, faça a seguinte substituição

$\begin{array}{lcl} \mathsf{x^2+1=u}&\quad\Rightarrow\quad& \mathsf{x^2=u-1}\\\\ &&\mathsf{2x\,dx=du} \end{array}$


e a integral fica

$\mathsf{=\displaystyle\frac{1}{2}\int\! (u-1)\cdot u^{3/2}\,du}\\\\\\ \mathsf{=\displaystyle\frac{1}{2}\int\! (u\cdot u^{3/2}-1\cdot u^{3/2})\,du}\\\\\\ \mathsf{=\displaystyle\frac{1}{2}\int\! (u^{1+(3/2)}-u^{3/2})\,du}\\\\\\ \mathsf{=\displaystyle\frac{1}{2}\int\! (u^{5/2}-u^{3/2})\,du}\\\\\\ \mathsf{=\displaystyle\frac{1}{2}\int\! u^{5/2}\,du-\frac{1}{2}\int\!u^{3/2}\,du}$


Aplique a regra para primitivar potências:

$\mathsf{=\dfrac{1}{2}\cdot \dfrac{u^{(5/2)+1}}{\frac{5}{2}+1}-\dfrac{1}{2}\cdot \dfrac{u^{(3/2)+1}}{\frac{3}{2}+1}+C}\\\\\\ \mathsf{=\dfrac{1}{2}\cdot \dfrac{u^{7/2}}{\frac{7}{2}}-\dfrac{1}{2}\cdot \dfrac{u^{5/2}}{\frac{5}{2}}+C}\\\\\\ \mathsf{=\dfrac{1}{\diagup\hspace{-7}2}\cdot \dfrac{\diagup\hspace{-7}2}{7}\,u^{7/2}-\dfrac{1}{\diagup\hspace{-7}2}\cdot \dfrac{\diagup\hspace{-7}2}{5}\,u^{5/2}+C}\\\\\\ \mathsf{=\dfrac{1}{7}\,u^{7/2}-\dfrac{1}{5}\,u^{5/2}+C}$


Por fim, você substitui de volta para a variável  x, obtendo

$\mathsf{=\dfrac{1}{7}\,(x^2+1)^{7/2}-\dfrac{1}{5}\,(x^2+1)^{5/2}+C}$


∴     $\mathsf{\displaystyle\int\! x^3\cdot (x^2+1)^{3/2}\,dx=\frac{1}{7}\,(x^2+1)^{7/2}-\frac{1}{5}\,(x^2+1)^{5/2}+C}$          ✔


Bons estudos! :-)


Tags:  integral indefinida substituição mudança de variável função irracional cálculo diferencial integral