Question

Resolva a integral indefinida:

$\mathsf{ \displaystyle \int \dfrac{\sqrt{(x-1)(2-x)}}{x} \, \, dx}}$

Resolução detalhada mostrando as técnicas utilizadas.

Answer 1

                       $\displaystyle I=\int \dfrac{\sqrt{(x-1)(2-x)}}{x} ~ dx\\\\\\ I=\int \dfrac{\sqrt{-2+3x-x^2}}{x} ~ dx\\ \\ \\ I=\int \dfrac{\sqrt{\frac{1}{4}-(x-\frac{3}{2})^2}}{x} ~ dx\\ \\ \\ \text{Sustituci\'on: }u=2\left(x-\frac{3}{2}\right)\to du=2~dx\to dx=\frac{du}{2}\\ \\ I=\int \dfrac{\sqrt{\frac{1}{4}-\frac{u^2}{4}}}{\frac{u+3}{2}} \cdot \frac{du}{2}$


$\displaystyle I=\frac{1}{2}\int \frac{\sqrt{1-u^2}}{u+3}~du\\ \\ \text{Sustituci\'on: }u=\sin t\to du=\cos t ~dt\\ \\ I=\frac{1}{2}\int \frac{\cos t}{\sin t+3}~\cos t ~dt\\ \\ \\ I=\frac{1}{2}\int \frac{\cos^2 t}{\sin t+3} ~dt\\ \\ \\ I=\frac{1}{2}\int \frac{1-\sin^2 t}{\sin t+3} ~dt\\ \\ \\ I=\frac{1}{2}\int 3-\sin t-\frac{8}{\sin t+3} ~dt\\ \\ \\ I=\frac{3}{2}t+\frac{1}{2}\cos t-4\int \frac{dt}{\sin t+3}$


$\displaystyle I=\frac{3}{2}t+\frac{1}{2}\cos t-4\int \frac{dt}{\cos (t-\frac{\pi}{2})+3} \\ \\ \\ I=\frac{3}{2}t+\frac{1}{2}\cos t-4\int \frac{dt}{\cos 2(\frac{t}{2}-\frac{\pi}{4})+3} \\ \\ \\ I=\frac{3}{2}t+\frac{1}{2}\cos t-4\int \frac{dt}{2\cos^ 2(\frac{t}{2}-\frac{\pi}{4})+2}$


$\displaystyle I=\frac{3}{2}t+\frac{1}{2}\cos t-2\int \frac{\sec^ 2(\frac{t}{2}-\frac{\pi}{4})}{\sec^ 2(\frac{t}{2}-\frac{\pi}{4})+1} ~dt\\ \\ \\ I=\frac{3}{2}t+\frac{1}{2}\cos t-2\int \frac{d\tan(\frac{t}{2}-\frac{\pi}{4})}{\tan^ 2(\frac{t}{2}-\frac{\pi}{4})+2}$


$\displaystyle I=\frac{3}{2}t+\frac{1}{2}\cos t-2\cdot\frac{1}{\sqrt2}\arctan\left[\frac{\tan(\frac{t}{2}-\frac{\pi}{4})}{\sqrt{2}}\right]+C\\ \\ \\ I=\frac{3}{2}t+\frac{1}{2}\cos t-\sqrt2\arctan\left(\frac{\tan t-\sec t}{\sqrt{2}}\right)+C\\ \\ \\ I=\frac{3}{2}\arcsin u+\frac{1}{2}\sqrt{1-u^2}-\sqrt2\arctan\left(-\frac{\sqrt{1-u^2}}{\sqrt{2}(u+1)}\right)+C\\ \\ \\ \\ I=\frac{3}{2}\arcsin (2x-3)+2\sqrt{(x-1)(2-x)}+\cdots\\ \hspace*{5cm}\cdots+\sqrt2\arctan\left(\sqrt{2}\sqrt{\frac{2-x}{x-1}}\right)+C$