Question

Como calcular a integral de (\/x +1) (x+\/x)

Answer 1

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Calcular a integral indefinida:

     $\mathsf{\displaystyle\int\!\sqrt{x+1}\cdot (x+\sqrt{x})\,dx}$


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:

     $\mathsf{\displaystyle\int\!\sqrt{x+1}\cdot (x+\sqrt{x})\,dx}\\\\\\ =\mathsf{\displaystyle\int\!\sqrt{x+1}\cdot \big[(x+1)-1+\sqrt{x}\big]dx}\\\\\\ =\mathsf{\displaystyle\int\!\big[\sqrt{x+1}\cdot (x+1)-\sqrt{x+1}\cdot 1+\sqrt{x+1}\cdot \sqrt{x}\big]dx}\\\\\\ =\mathsf{\displaystyle\int\!\big[(x+1)^{1/2}\cdot (x+1)-(x+1)^{1/2}+\sqrt{x+1}\cdot \sqrt{x}\big]dx}\\\\\\ =\mathsf{\displaystyle\int\!\big[(x+1)^{(1/2)+1}-(x+1)^{1/2}+\sqrt{x+1}\cdot \sqrt{x}\big]dx}\\\\\\ =\mathsf{\displaystyle\int\!\big[(x+1)^{3/2}-(x+1)^{1/2}+\sqrt{x+1}\cdot \sqrt{x}\big]dx}$


Separe como uma soma de duas integrais:

     $=\mathsf{\displaystyle\int\!\big[(x+1)^{3/2}-(x+1)^{1/2}\big]dx+\int\!\sqrt{x+1}\cdot \sqrt{x}\,dx}\\\\\\ =\mathsf{I_1+I_2\qquad\quad(i)}$

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Calculando  as integrais separadamente:

•   $\mathsf{I_1=\displaystyle\int\!\big[(x+1)^{3/2}-(x+1)^{1/2}\big]dx}$


Substituição:

     $\mathsf{x+1=u\quad\Rightarrow\quad dx=du}$


e a integral  I₁  fica

     $=\mathsf{\displaystyle\int\!\big[u^{3/2}-u^{1/2}\big]du}\\\\\\ =\mathsf{\dfrac{u^{(3/2)+1}}{\frac{3}{2}+1}-\dfrac{u^{(1/2)+1}}{\frac{1}{2}+1}+C_1}\\\\\\ =\mathsf{\dfrac{u^{5/2}}{\frac{5}{2}}-\dfrac{u^{3/2}}{\frac{3}{2}}+C_1}\\\\\\ =\mathsf{\dfrac{2}{5}\,u^{5/2}-\dfrac{2}{3}\,u^{3/2}+C_1}\\\\\\ =\mathsf{\dfrac{2}{5}\,(x+1)^{5/2}-\dfrac{2}{3}\,(x+1)^{3/2}+C_1\qquad\quad(ii)}$

sendo  C₁  uma constante.

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•   $\mathsf{I_2=\displaystyle\int\!\sqrt{x+1}\cdot \sqrt{x}\,dx}$

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


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

     $=\mathsf{\displaystyle\int\!\sqrt{\left(x^2+x+\frac{1}{4}\right)-\frac{1}{4}}\,dx}\\\\\\ =\mathsf{\displaystyle\int\!\sqrt{\left(x+\frac{1}{2}\right)^{\!\!2}-\left(\frac{1}{2}\right)^{\!\!2}}\,dx}\\\\\\ =\mathsf{\displaystyle\int\!\sqrt{v^2-a^2}\,dv\qquad\quad(iii)}$


onde   $\left\{\! \begin{array}{l} \mathsf{v=x+\dfrac{1}{2}\quad\Rightarrow\quad dv=dx}\\\\ \mathsf{a=\dfrac{1}{2}} \end{array} \right.$


A integral na forma apresentada em  (iii)  é tabelada  

    $\mathsf{\displaystyle\int\!\sqrt{v^2-a^2}\,dv=\frac{1}{2}\,v\sqrt{v^2-a^2}-\frac{a^2}{2}\,ln\big(\sqrt{v^2-a^2}+v\big)}$


onde  a  é uma constante positiva;  de modo que a integral  I₂  fica

$=\mathsf{\dfrac{1}{2}\!\left(x+\dfrac{1}{2}\right)\!\sqrt{\left(x+\dfrac{1}{2}\right)^{\!\!2}-\left(\dfrac{1}{2}\right)^{\!\!2}}-\dfrac{(\frac{1}{2})^2}{2}\,ln\!\left[\sqrt{\left(x+\dfrac{1}{2}\right)^{\!\!2}-\left(\dfrac{1}{2}\right)^{\!\!2}}+x+\dfrac{1}{2}\right]+C_2}\\\\\\ =\mathsf{\dfrac{1}{2}\!\left(x+\dfrac{1}{2}\right)\!\sqrt{x^2+x}-\dfrac{~\frac{1}{4}~}{2}\,ln\!\left[\sqrt{x^2+x}+x+\dfrac{1}{2}\right]+C_2}\\\\\\ =\mathsf{\dfrac{1}{2}\!\left(x+\dfrac{1}{2}\right)\!\sqrt{(x+1)\cdot x}-\dfrac{1}{8}\,ln\!\left[\sqrt{(x+1)\cdot x}+x+\dfrac{1}{2}\right]+C_2}\\\\\\ =\mathsf{\dfrac{1}{2}\!\left(x+\dfrac{1}{2}\right)\!\sqrt{x+1}\cdot \sqrt{x}-\dfrac{1}{8}\,ln\!\left[\sqrt{x+1}\cdot \sqrt{x}+x+\dfrac{1}{2}\right]+C_2\qquad\quad(iv)}$

sendo  C₂  uma constante.

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Logo, a integral pedida nesta tarefa é

$=\mathsf{\displaystyle\int\!\sqrt{x+1}\cdot (x+\sqrt{x})\,dx}\\\\\\ =\mathsf{I_1+I_2}$


$=\mathsf{\dfrac{2}{5}\,(x+1)^{5/2}-\dfrac{2}{3}\,(x+1)^{3/2}+\dfrac{1}{2}\!\left(x+\dfrac{1}{2}\right)\!\sqrt{x+1}\cdot \sqrt{x}-\dfrac{1}{8}\,ln\!\left[\sqrt{x+1}\cdot \sqrt{x}+x+\dfrac{1}{2}\right]+C}$


Bons estudos! :-)