Como calcular a integral de (\/x +1) (x+\/x)
Lukyo:
No primeiro parênteses, o que está dentro da raiz? é só o x, ou é raiz quadrada de (x + 1)? "+ 1" está na raiz quadrada também ou é só o x?
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Calcular a integral indefinida:

O domínio do integrando é determinado pelas condições
x + 1 ≥ 0 e x ≥ 0
x ≥ – 1 e x ≥ 0
x ≥ 0
Reescreva o primeiro x como (x + 1) – 1 para facilitar a próxima substituição, e aplique a distributiva:
![\mathsf{\displaystyle\int\!\sqrt{x+1}\cdot (x+\sqrt{x})\,dx}\\\\\\ =\mathsf{\displaystyle\int\!\sqrt{x+1}\cdot \big[(x+1)-1+\sqrt{x}\big]dx}\\\\\\ =\mathsf{\displaystyle\int\!\big[\sqrt{x+1}\cdot (x+1)-\sqrt{x+1}\cdot 1+\sqrt{x+1}\cdot \sqrt{x}\big]dx}\\\\\\ =\mathsf{\displaystyle\int\!\big[(x+1)^{1/2}\cdot (x+1)-(x+1)^{1/2}+\sqrt{x+1}\cdot \sqrt{x}\big]dx}\\\\\\ =\mathsf{\displaystyle\int\!\big[(x+1)^{(1/2)+1}-(x+1)^{1/2}+\sqrt{x+1}\cdot \sqrt{x}\big]dx}\\\\\\ =\mathsf{\displaystyle\int\!\big[(x+1)^{3/2}-(x+1)^{1/2}+\sqrt{x+1}\cdot \sqrt{x}\big]dx} \mathsf{\displaystyle\int\!\sqrt{x+1}\cdot (x+\sqrt{x})\,dx}\\\\\\ =\mathsf{\displaystyle\int\!\sqrt{x+1}\cdot \big[(x+1)-1+\sqrt{x}\big]dx}\\\\\\ =\mathsf{\displaystyle\int\!\big[\sqrt{x+1}\cdot (x+1)-\sqrt{x+1}\cdot 1+\sqrt{x+1}\cdot \sqrt{x}\big]dx}\\\\\\ =\mathsf{\displaystyle\int\!\big[(x+1)^{1/2}\cdot (x+1)-(x+1)^{1/2}+\sqrt{x+1}\cdot \sqrt{x}\big]dx}\\\\\\ =\mathsf{\displaystyle\int\!\big[(x+1)^{(1/2)+1}-(x+1)^{1/2}+\sqrt{x+1}\cdot \sqrt{x}\big]dx}\\\\\\ =\mathsf{\displaystyle\int\!\big[(x+1)^{3/2}-(x+1)^{1/2}+\sqrt{x+1}\cdot \sqrt{x}\big]dx}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%5Csqrt%7Bx%2B1%7D%5Ccdot+%28x%2B%5Csqrt%7Bx%7D%29%5C%2Cdx%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%5Csqrt%7Bx%2B1%7D%5Ccdot+%5Cbig%5B%28x%2B1%29-1%2B%5Csqrt%7Bx%7D%5Cbig%5Ddx%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%5Cbig%5B%5Csqrt%7Bx%2B1%7D%5Ccdot+%28x%2B1%29-%5Csqrt%7Bx%2B1%7D%5Ccdot+1%2B%5Csqrt%7Bx%2B1%7D%5Ccdot+%5Csqrt%7Bx%7D%5Cbig%5Ddx%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%5Cbig%5B%28x%2B1%29%5E%7B1%2F2%7D%5Ccdot+%28x%2B1%29-%28x%2B1%29%5E%7B1%2F2%7D%2B%5Csqrt%7Bx%2B1%7D%5Ccdot+%5Csqrt%7Bx%7D%5Cbig%5Ddx%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%5Cbig%5B%28x%2B1%29%5E%7B%281%2F2%29%2B1%7D-%28x%2B1%29%5E%7B1%2F2%7D%2B%5Csqrt%7Bx%2B1%7D%5Ccdot+%5Csqrt%7Bx%7D%5Cbig%5Ddx%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%5Cbig%5B%28x%2B1%29%5E%7B3%2F2%7D-%28x%2B1%29%5E%7B1%2F2%7D%2B%5Csqrt%7Bx%2B1%7D%5Ccdot+%5Csqrt%7Bx%7D%5Cbig%5Ddx%7D)
Separe como uma soma de duas integrais:
![=\mathsf{\displaystyle\int\!\big[(x+1)^{3/2}-(x+1)^{1/2}\big]dx+\int\!\sqrt{x+1}\cdot \sqrt{x}\,dx}\\\\\\ =\mathsf{I_1+I_2\qquad\quad(i)} =\mathsf{\displaystyle\int\!\big[(x+1)^{3/2}-(x+1)^{1/2}\big]dx+\int\!\sqrt{x+1}\cdot \sqrt{x}\,dx}\\\\\\ =\mathsf{I_1+I_2\qquad\quad(i)}](https://tex.z-dn.net/?f=%3D%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%5Cbig%5B%28x%2B1%29%5E%7B3%2F2%7D-%28x%2B1%29%5E%7B1%2F2%7D%5Cbig%5Ddx%2B%5Cint%5C%21%5Csqrt%7Bx%2B1%7D%5Ccdot+%5Csqrt%7Bx%7D%5C%2Cdx%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7BI_1%2BI_2%5Cqquad%5Cquad%28i%29%7D)
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Calculando as integrais separadamente:
•![\mathsf{I_1=\displaystyle\int\!\big[(x+1)^{3/2}-(x+1)^{1/2}\big]dx} \mathsf{I_1=\displaystyle\int\!\big[(x+1)^{3/2}-(x+1)^{1/2}\big]dx}](https://tex.z-dn.net/?f=%5Cmathsf%7BI_1%3D%5Cdisplaystyle%5Cint%5C%21%5Cbig%5B%28x%2B1%29%5E%7B3%2F2%7D-%28x%2B1%29%5E%7B1%2F2%7D%5Cbig%5Ddx%7D)
Substituição:

e a integral I₁ fica
![=\mathsf{\displaystyle\int\!\big[u^{3/2}-u^{1/2}\big]du}\\\\\\ =\mathsf{\dfrac{u^{(3/2)+1}}{\frac{3}{2}+1}-\dfrac{u^{(1/2)+1}}{\frac{1}{2}+1}+C_1}\\\\\\ =\mathsf{\dfrac{u^{5/2}}{\frac{5}{2}}-\dfrac{u^{3/2}}{\frac{3}{2}}+C_1}\\\\\\ =\mathsf{\dfrac{2}{5}\,u^{5/2}-\dfrac{2}{3}\,u^{3/2}+C_1}\\\\\\ =\mathsf{\dfrac{2}{5}\,(x+1)^{5/2}-\dfrac{2}{3}\,(x+1)^{3/2}+C_1\qquad\quad(ii)} =\mathsf{\displaystyle\int\!\big[u^{3/2}-u^{1/2}\big]du}\\\\\\ =\mathsf{\dfrac{u^{(3/2)+1}}{\frac{3}{2}+1}-\dfrac{u^{(1/2)+1}}{\frac{1}{2}+1}+C_1}\\\\\\ =\mathsf{\dfrac{u^{5/2}}{\frac{5}{2}}-\dfrac{u^{3/2}}{\frac{3}{2}}+C_1}\\\\\\ =\mathsf{\dfrac{2}{5}\,u^{5/2}-\dfrac{2}{3}\,u^{3/2}+C_1}\\\\\\ =\mathsf{\dfrac{2}{5}\,(x+1)^{5/2}-\dfrac{2}{3}\,(x+1)^{3/2}+C_1\qquad\quad(ii)}](https://tex.z-dn.net/?f=%3D%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%5Cbig%5Bu%5E%7B3%2F2%7D-u%5E%7B1%2F2%7D%5Cbig%5Ddu%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdfrac%7Bu%5E%7B%283%2F2%29%2B1%7D%7D%7B%5Cfrac%7B3%7D%7B2%7D%2B1%7D-%5Cdfrac%7Bu%5E%7B%281%2F2%29%2B1%7D%7D%7B%5Cfrac%7B1%7D%7B2%7D%2B1%7D%2BC_1%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdfrac%7Bu%5E%7B5%2F2%7D%7D%7B%5Cfrac%7B5%7D%7B2%7D%7D-%5Cdfrac%7Bu%5E%7B3%2F2%7D%7D%7B%5Cfrac%7B3%7D%7B2%7D%7D%2BC_1%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdfrac%7B2%7D%7B5%7D%5C%2Cu%5E%7B5%2F2%7D-%5Cdfrac%7B2%7D%7B3%7D%5C%2Cu%5E%7B3%2F2%7D%2BC_1%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdfrac%7B2%7D%7B5%7D%5C%2C%28x%2B1%29%5E%7B5%2F2%7D-%5Cdfrac%7B2%7D%7B3%7D%5C%2C%28x%2B1%29%5E%7B3%2F2%7D%2BC_1%5Cqquad%5Cquad%28ii%29%7D)
sendo C₁ uma constante.
—————
•

Some e subtraia 1/4 para completar o trinômio quadrado perfeito no radicando:

onde
A integral na forma apresentada em (iii) é tabelada

onde a é uma constante positiva; de modo que a integral I₂ fica
![=\mathsf{\dfrac{1}{2}\!\left(x+\dfrac{1}{2}\right)\!\sqrt{\left(x+\dfrac{1}{2}\right)^{\!\!2}-\left(\dfrac{1}{2}\right)^{\!\!2}}-\dfrac{(\frac{1}{2})^2}{2}\,ln\!\left[\sqrt{\left(x+\dfrac{1}{2}\right)^{\!\!2}-\left(\dfrac{1}{2}\right)^{\!\!2}}+x+\dfrac{1}{2}\right]+C_2}\\\\\\ =\mathsf{\dfrac{1}{2}\!\left(x+\dfrac{1}{2}\right)\!\sqrt{x^2+x}-\dfrac{~\frac{1}{4}~}{2}\,ln\!\left[\sqrt{x^2+x}+x+\dfrac{1}{2}\right]+C_2}\\\\\\ =\mathsf{\dfrac{1}{2}\!\left(x+\dfrac{1}{2}\right)\!\sqrt{(x+1)\cdot x}-\dfrac{1}{8}\,ln\!\left[\sqrt{(x+1)\cdot x}+x+\dfrac{1}{2}\right]+C_2}\\\\\\ =\mathsf{\dfrac{1}{2}\!\left(x+\dfrac{1}{2}\right)\!\sqrt{x+1}\cdot \sqrt{x}-\dfrac{1}{8}\,ln\!\left[\sqrt{x+1}\cdot \sqrt{x}+x+\dfrac{1}{2}\right]+C_2\qquad\quad(iv)} =\mathsf{\dfrac{1}{2}\!\left(x+\dfrac{1}{2}\right)\!\sqrt{\left(x+\dfrac{1}{2}\right)^{\!\!2}-\left(\dfrac{1}{2}\right)^{\!\!2}}-\dfrac{(\frac{1}{2})^2}{2}\,ln\!\left[\sqrt{\left(x+\dfrac{1}{2}\right)^{\!\!2}-\left(\dfrac{1}{2}\right)^{\!\!2}}+x+\dfrac{1}{2}\right]+C_2}\\\\\\ =\mathsf{\dfrac{1}{2}\!\left(x+\dfrac{1}{2}\right)\!\sqrt{x^2+x}-\dfrac{~\frac{1}{4}~}{2}\,ln\!\left[\sqrt{x^2+x}+x+\dfrac{1}{2}\right]+C_2}\\\\\\ =\mathsf{\dfrac{1}{2}\!\left(x+\dfrac{1}{2}\right)\!\sqrt{(x+1)\cdot x}-\dfrac{1}{8}\,ln\!\left[\sqrt{(x+1)\cdot x}+x+\dfrac{1}{2}\right]+C_2}\\\\\\ =\mathsf{\dfrac{1}{2}\!\left(x+\dfrac{1}{2}\right)\!\sqrt{x+1}\cdot \sqrt{x}-\dfrac{1}{8}\,ln\!\left[\sqrt{x+1}\cdot \sqrt{x}+x+\dfrac{1}{2}\right]+C_2\qquad\quad(iv)}](https://tex.z-dn.net/?f=%3D%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5C%21%5Cleft%28x%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5C%21%5Csqrt%7B%5Cleft%28x%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B%5C%21%5C%212%7D-%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B%5C%21%5C%212%7D%7D-%5Cdfrac%7B%28%5Cfrac%7B1%7D%7B2%7D%29%5E2%7D%7B2%7D%5C%2Cln%5C%21%5Cleft%5B%5Csqrt%7B%5Cleft%28x%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B%5C%21%5C%212%7D-%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B%5C%21%5C%212%7D%7D%2Bx%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%5D%2BC_2%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5C%21%5Cleft%28x%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5C%21%5Csqrt%7Bx%5E2%2Bx%7D-%5Cdfrac%7B%7E%5Cfrac%7B1%7D%7B4%7D%7E%7D%7B2%7D%5C%2Cln%5C%21%5Cleft%5B%5Csqrt%7Bx%5E2%2Bx%7D%2Bx%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%5D%2BC_2%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5C%21%5Cleft%28x%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5C%21%5Csqrt%7B%28x%2B1%29%5Ccdot+x%7D-%5Cdfrac%7B1%7D%7B8%7D%5C%2Cln%5C%21%5Cleft%5B%5Csqrt%7B%28x%2B1%29%5Ccdot+x%7D%2Bx%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%5D%2BC_2%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5C%21%5Cleft%28x%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5C%21%5Csqrt%7Bx%2B1%7D%5Ccdot+%5Csqrt%7Bx%7D-%5Cdfrac%7B1%7D%7B8%7D%5C%2Cln%5C%21%5Cleft%5B%5Csqrt%7Bx%2B1%7D%5Ccdot+%5Csqrt%7Bx%7D%2Bx%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%5D%2BC_2%5Cqquad%5Cquad%28iv%29%7D)
sendo C₂ uma constante.
—————
Logo, a integral pedida nesta tarefa é

![=\mathsf{\dfrac{2}{5}\,(x+1)^{5/2}-\dfrac{2}{3}\,(x+1)^{3/2}+\dfrac{1}{2}\!\left(x+\dfrac{1}{2}\right)\!\sqrt{x+1}\cdot \sqrt{x}-\dfrac{1}{8}\,ln\!\left[\sqrt{x+1}\cdot \sqrt{x}+x+\dfrac{1}{2}\right]+C} =\mathsf{\dfrac{2}{5}\,(x+1)^{5/2}-\dfrac{2}{3}\,(x+1)^{3/2}+\dfrac{1}{2}\!\left(x+\dfrac{1}{2}\right)\!\sqrt{x+1}\cdot \sqrt{x}-\dfrac{1}{8}\,ln\!\left[\sqrt{x+1}\cdot \sqrt{x}+x+\dfrac{1}{2}\right]+C}](https://tex.z-dn.net/?f=%3D%5Cmathsf%7B%5Cdfrac%7B2%7D%7B5%7D%5C%2C%28x%2B1%29%5E%7B5%2F2%7D-%5Cdfrac%7B2%7D%7B3%7D%5C%2C%28x%2B1%29%5E%7B3%2F2%7D%2B%5Cdfrac%7B1%7D%7B2%7D%5C%21%5Cleft%28x%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5C%21%5Csqrt%7Bx%2B1%7D%5Ccdot+%5Csqrt%7Bx%7D-%5Cdfrac%7B1%7D%7B8%7D%5C%2Cln%5C%21%5Cleft%5B%5Csqrt%7Bx%2B1%7D%5Ccdot+%5Csqrt%7Bx%7D%2Bx%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%5D%2BC%7D)
Bons estudos! :-)
——————————
Calcular a integral indefinida:
O domínio do integrando é determinado pelas condições
x + 1 ≥ 0 e x ≥ 0
x ≥ – 1 e x ≥ 0
x ≥ 0
Reescreva o primeiro x como (x + 1) – 1 para facilitar a próxima substituição, e aplique a distributiva:
Separe como uma soma de duas integrais:
—————
Calculando as integrais separadamente:
•
Substituição:
e a integral I₁ fica
sendo C₁ uma constante.
—————
•
Some e subtraia 1/4 para completar o trinômio quadrado perfeito no radicando:
onde
A integral na forma apresentada em (iii) é tabelada
onde a é uma constante positiva; de modo que a integral I₂ fica
sendo C₂ uma constante.
—————
Logo, a integral pedida nesta tarefa é
Bons estudos! :-)
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