Question

Calcule a integral indefinida

g) $\int\ {}(3 \sqrt{x}-2)(3 \sqrt{x+2} \, dx$

Answer 1

Olá Lucas!

Antes de integrar iremos utilizar a diferença do quadrado perfeito:

$(a-b)(a+b) = a^2-b^2$

Nesse caso,

$\left \{ {{a=3\sqrt{x} } \atop {b=2}} \right.$

Logo,

$\\ (3 \sqrt{x} -2)(3 \sqrt{x} +2) = (3 \sqrt{x} )^2-2^2 \\ \\ (3 \sqrt{x} -2)(3 \sqrt{x} +2) = 9x -4$

Então,

$\\ (3 \sqrt{x} -2)(3 \sqrt{x} +2) = 9x-4 \\ \\ ou \\ \\ (3 \sqrt{x} -2)(3 \sqrt{x} +2) = -9x-4$
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Nossa  integral fica  fica:

$\int\limits 9x-4{} \, dx =$

Usando integração por potência e constante:

$\\ \int\limits {x^n} \, dx = \frac{x^n^+^1}{n+1} \\ \\ \int\limits {k} \, dx = kx$

Então,

$\\ = \frac{9x^1^+^1}{1+1} -4x \\ \\ = \frac{9x^2}{2} -4x$

Lembrando da constante "k"

$\\ = \frac{9x^2}{2} -4x+K$
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Answer 2

$I=\displaystyle\int\!(3\sqrt{x}-2)(3\sqrt{x+2})\,dx\\\\\\ =\int\!\big(3\sqrt{x}\cdot 3\sqrt{x+2}-2\cdot 3\sqrt{x+2}\big)\,dx\\\\\\ =\int\!\big(9\sqrt{x(x+2)}-6\sqrt{x+2}\big)\,dx\\\\\\ =\int\!\big(9\sqrt{x^2+2x}-6\sqrt{x+2}\big)\,dx\\\\\\ =\int\!\big(9\sqrt{x^2+2x+1-1}-6\sqrt{x+2}\big)\,dx\\\\\\ =\int\!\big(9\sqrt{(x+1)^2-1}-6\sqrt{x+2}\big)\,dx\\\\\\ =9\int\!\sqrt{(x+1)^2-1}-6\int\!\sqrt{x+2}\,dx~~~~~~\mathbf{(i)}$


Vamos avaliar as integrais separadamente:

$\displaystyle I_1=\int\!\sqrt{(x+1)^2-1}\,dx$


Substituição trigonométrica:

$x+1=\sec t~~\Rightarrow~~\left\{\! \begin{array}{l} dx=\sec t\,\mathrm{tg\,}t\,dt\\\\ 0\le t<\dfrac{\pi}{2} \end{array} \right.\\\\\\ (x+1)^2-1=\sec^2 t-1\\\\ (x+1)^2-1=\mathrm{tg^2\,}t\\\\ \sqrt{(x+1)^2-1}=\sqrt{\mathrm{tg^2\,}t}\\\\ \sqrt{(x+1)^2-1}=\left|\mathrm{tg\,}t\right|\\\\ \sqrt{(x+1)^2-1}=\mathrm{tg\,}t$


( pois $t$ está no 1º quadrante )


Substituindo, a integral $I_1$ fica

$\displaystyle I_1=\int\!\mathrm{tg\,}t\cdot \sec t\,\mathrm{tg\,}t\,dt\\\\\\ =\int\!\mathrm{tg^2\,}t\cdot \sec t\,dt\\\\\\ =\int\!(\sec^2 t-1)\cdot \sec t\,dt\\\\\\ =\int\!(\sec^3 t\,dt-\sec t)\,dt\\\\\\ =\int\!\sec^3 t\,dt-\int\!\sec t\,dt\\\\\\ =\underbrace{\int\!\sec^2 t\cdot \sec t\,dt}_{I_2}-\int\!\sec t\,dt~~~~~~\mathbf{(ii)}$


Resolvemos $I_2$ por partes:

$\begin{array}{lcl} u=\sec t&~\Rightarrow~&du=\sec t\,\mathrm{tg\,}t\,dt\\\\ dv=\sec^2 t\,dt&~\Leftarrow~&v=\mathrm{tg\,}t \end{array}\\\\\\ \displaystyle\int\!u\,dv=uv-\int\!v\,du\\\\\\ \int\!\sec^2 t\cdot \sec t\,dt=\sec t\,\mathrm{tg\,}t-\int\!\sec t\cdot \mathrm{tg^2\,}t\,dt\\\\\\ I_2=\sec t\,\mathrm{tg\,}t-I_1~~~~~~\mathbf{(iii)}$


Substituindo $\mathbf{(iii)}$ em $\mathbf{(ii)},$ ficamos com

$\displaystyle I_1=I_2-\int\!\sec t\,dt\\\\\\ I_1=(\sec t\,\mathrm{tg\,}t-I_1)-\int\!\sec t\,dt\\\\\\ 2I_1=\sec t\,\mathrm{tg\,}t-\int\!\sec t\,dt\\\\\\ I_1=\dfrac{1}{2}\sec t\,\mathrm{tg\,}t-\dfrac{1}{2}\int\!\sec t\,dt\\\\\\ I_1=\dfrac{1}{2}\sec t\,\mathrm{tg\,}t-\dfrac{1}{2}\,\mathrm{\ell n}\left|\sec t+\mathrm{tg\,}t\right|\\\\\\ \boxed{\begin{array}{c}I_1=\dfrac{1}{2}\,(x+1)\sqrt{x(x+2)}-\dfrac{1}{2}\,\mathrm{\ell n}\left|(x+1)+\sqrt{x(x+2)}\right| \end{array}}$

______________

$\displaystyle I_3=\int\!\sqrt{x+2}\,dx\\\\\\ =\int\! (x+2)^{1/2}\,dx\\\\\\ =\dfrac{(x+2)^{(1/2)+1}}{\frac{1}{2}+1}\\\\\\ =\dfrac{(x+2)^{3/2}}{\frac{3}{2}}\\\\\\ \therefore~~\boxed{\begin{array}{c}I_3=\dfrac{2}{3}\,(x+2)^{3/2}\end{array}}$

______________

Voltando a $\mathbf{(i)},$ a integral pedida é

$I=9I_1-6I_3\\\\\\ I=9\cdot \left(\dfrac{1}{2}\,(x+1)\sqrt{x(x+2)}-\dfrac{1}{2}\,\mathrm{\ell n}\left|(x+1)+\sqrt{x(x+2)}\right| \right )-6\cdot \dfrac{2}{3}\,(x+2)^{3/2}+C\\\\\\ \therefore~~\boxed{\begin{array}{c}I=\dfrac{9}{2}\,(x+1)\sqrt{x(x+2)}-\dfrac{9}{2}\,\mathrm{\ell n}\left|(x+1)+\sqrt{x(x+2)}\right|-4(x+2)^{3/2}+C \end{array}}$


Bons estudos! :-)