Como calcular a integral de (\/x +1) (x+\/x).... Pergunta de ideia deMhary11

December 2019 1 111 Report

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

Lista de comentários

Lukyo

Verified answer

Caso esteja pelo app, e tenha problemas para visualizar esta resposta, experimente abrir pelo navegador: brainly.com.br/tarefa/10287481

——————————

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:

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

—————

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.

—————

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

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

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.

—————

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! :-)

2 votes Thanks 2