Question

INTEGRAL DEFINIDA DE 1 A 4

[x2 - 2x - 3raiz de x - 1/x2] dx

Answer 1

$I=\displaystyle\int\limits_{1}^{4}{\left(x^{2}-2x-\dfrac{3\sqrt{x-1}}{x^{2}} \right )dx}\\ \\ \\ =\int\limits_{1}^{4}{(x^{2}-2x)\,dx}-3\int\limits_{1}^{4}{\dfrac{\sqrt{x-1}}{x^{2}}\,dx}\\ \\ \\ =\left.\left(\dfrac{x^{3}}{3}-x^{2} \right )\right|_{1}^{4}-3\int\limits_{1}^{4}{\dfrac{\sqrt{x-1}}{x^{2}}\,dx}\\ \\ \\ =\left[\left(\dfrac{4^{3}}{3}-4^{2} \right )-\left(\dfrac{1^{3}}{3}-1^{2} \right ) \right ]-3\int\limits_{1}^{4}{\dfrac{\sqrt{x-1}}{x^{2}}\,dx}$

$=\displaystyle\left[\dfrac{64}{3}-16-\dfrac{1}{3}+1 \right ]-3\int\limits_{1}^{4}{\dfrac{\sqrt{x-1}}{x^{2}}\,dx}\\ \\ \\ =\left[\dfrac{64-1}{3}-15 \right ]-3\int\limits_{1}^{4}{\dfrac{\sqrt{x-1}}{x^{2}}\,dx}\\ \\ \\ =\left[21-15 \right ]-3\int\limits_{1}^{4}{\dfrac{\sqrt{x-1}}{x^{2}}\,dx}\\ \\ \\ =6-3\underbrace{\int\limits_{1}^{4}{\dfrac{\sqrt{x-1}}{x^{2}}\,dx}}_{I_{1}}\;\;\;\;\;\;\;\mathbf{(i)}$

$\bullet\;\;$ Calculando a integral $I_{1}:$

$I_{1}=\displaystyle\int\limits_{1}^{4}{\dfrac{\sqrt{x-1}}{x^{2}}\,dx}$


Façamos a seguinte mudança de variável:

$u=\sqrt{x-1}~\Rightarrow~u^{2}=x-1~\Rightarrow~\left\{ \begin{array}{l} x=u^{2}+1\\ dx=2u\,du \end{array} \right.$


Mudando os extremos de integração:

$\text{Quando }x=1~\Rightarrow~u=0\\ \\ \text{Quando }x=4~\Rightarrow~u=\sqrt{3}$


Substituindo em $I_{1},$ temos

$I_{1}=\displaystyle\int\limits_{0}^{\sqrt{3}}{\dfrac{u}{(u^{2}+1)^{2}}\cdot 2u\,du}\\ \\ \\ =2\int\limits_{0}^{\sqrt{3}}{\dfrac{u^{2}}{(u^{2}+1)^{2}}\,du}\\ \\ \\ =2\int\limits_{0}^{\sqrt{3}}{\dfrac{u^{2}+1-1}{(u^{2}+1)^{2}}\,du}\\ \\ \\ =2\int\limits_{0}^{\sqrt{3}}{\dfrac{u^{2}+1}{(u^{2}+1)^{2}}\,du}-2\int\limits_{0}^{\sqrt{3}}{\dfrac{1 }{(u^{2}+1)^{2}}\,du}\\ \\ \\ =2\int\limits_{0}^{\sqrt{3}}{\dfrac{1}{u^{2}+1}\,du}-2\int\limits_{0}^{\sqrt{3}}{\dfrac{1 }{(u^{2}+1)^{2}}\,du}\\ \\ \\ =2\,\mathrm{arctg\,}u|_{0}^{\sqrt{3}}-2\int\limits_{0}^{\sqrt{3}}{\dfrac{1 }{(u^{2}+1)^{2}}\,du}\\ \\ \\ =2\cdot \left(\mathrm{arctg\,}\sqrt{3}-\mathrm{arctg\,}0 \right )-2\int\limits_{0}^{\sqrt{3}}{\dfrac{1 }{(u^{2}+1)^{2}}\,du}$

$=\displaystyle\dfrac{2\pi}{3}-2\underbrace{\int\limits_{0}^{\sqrt{3}}{\dfrac{1 }{(u^{2}+1)^{2}}\,du}}_{I_{2}}\;\;\;\;\;\;\;\mathbf{(ii)}$


$\bullet\;\;$ Calcularemos $I_{2}$ por substituição trigonométrica

(Poderia ser por frações parciais, mas teria que fatorar um polinômio e grau 4... quero poupar trabalho :-D )

$I_{2}=\displaystyle\int\limits_{0}^{\sqrt{3}}{\dfrac{1}{(u^{2}+1)^{2}}\,du}$


Façamos a seguinte substituição:

$u=\mathrm{tg\,}t~\Rightarrow~t=\mathrm{arctg\,}u~\Rightarrow~\left\{ \begin{array}{l} du=\sec^{2}t\,dt\\ \\ u^{2}+1=\sec^{2}t \end{array} \right.$


Mudando os extremos de integração:

$\text{Quando }u=0~\Rightarrow~t=0\\ \\ \text{Quando }u=\sqrt{3}~\Rightarrow~t=\dfrac{\pi}{3}$


Substituindo em $I_{2},$ temos

$I_{2}=\displaystyle\int\limits_{0}^{\pi/3}{\dfrac{1}{(\sec^{2}t)^{2}}\cdot \sec^{2}t\,dt}\\ \\ \\ =\int\limits_{0}^{\pi/3}{\dfrac{\sec^{2}t}{\sec^{4}t}\,dt}\\ \\ \\ =\int\limits_{0}^{\pi/3}{\dfrac{1}{\sec^{2}t}\,dt}\\ \\ \\ =\int\limits_{0}^{\pi/3}{\cos^{2}t\,dt}\\ \\ \\ =\int\limits_{0}^{\pi/3}{\left(\dfrac{1}{2}+\dfrac{1}{2}\cos 2t \right )dt}\\ \\ \\ =\left.\left(\dfrac{t}{2}+\dfrac{1}{4}\,\mathrm{sen\,} 2t \right )\right|_{0}^{\pi/3}$

$=\left[\left(\dfrac{(\frac{\pi}{3})}{2}+\dfrac{1}{4}\,\mathrm{sen\,} \dfrac{2\pi}{3} \right )-\left(\dfrac{0}{2}+\dfrac{1}{4}\,\mathrm{sen\,} 0 \right ) \right ]\\ \\ \\ =\dfrac{\pi}{6}+\dfrac{1}{4}\cdot \dfrac{\sqrt{3}}{2}\\ \\ \\ =\dfrac{\pi}{6}+\dfrac{\sqrt{3}}{8}.$


$\bullet\;\;$ Substituindo em $\mathbf{(ii)},$ temos

$I_{1}=\dfrac{2\pi}{3}-2\cdot \left(\dfrac{\pi}{6}+\dfrac{\sqrt{3}}{8} \right )\\ \\ \\ =\dfrac{2\pi}{3}-\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{4}\\ \\ \\ =\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{4}.$


$\bullet\;\;$ Substituindo em $\mathbf{(i)},$ finalmente obtemos

$I=6-3\cdot \left(\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{4} \right )\\ \\ \\ =6-\pi+\dfrac{3\sqrt{3}}{4}\\ \\ \\ \\ \Rightarrow~\boxed{\begin{array}{l} \displaystyle\int\limits_{1}^{4}{\left(x^{2}-2x-\dfrac{3\sqrt{x-1}}{x^{2}} \right )dx}=6-\pi+\dfrac{3\sqrt{3}}{4} \end{array}}$