# Question #7f1cc

Nov 23, 2017

$\int \sqrt{\frac{x}{2 - x}} \cdot \mathrm{dx} = 2 \arctan \sqrt{\frac{x}{2 - x}} - \sqrt{2 x - {x}^{2}} + C$

#### Explanation:

$\int \sqrt{\frac{x}{2 - x}} \cdot \mathrm{dx}$

After using $y = \sqrt{\frac{x}{2 - x}}$ substitution,

${y}^{2} = \frac{x}{2 - x}$

$x = {y}^{2} \cdot \left(2 - x\right)$

$x = 2 {y}^{2} - x {y}^{2}$

$\left({y}^{2} + 1\right) \cdot x = 2 {y}^{2}$

$x = \frac{2 {y}^{2}}{{y}^{2} + 1}$

$\mathrm{dx} = \frac{\left[4 y \cdot \left({y}^{2} + 1\right) - 2 y \cdot 2 {y}^{2}\right] \cdot \mathrm{dy}}{{y}^{2} + 1} ^ 2$

=$\frac{4 y \cdot \mathrm{dy}}{{y}^{2} + 1} ^ 2$

After it, this integral became

$\int y \cdot \frac{4 y \cdot \mathrm{dy}}{{y}^{2} + 1} ^ 2$

=$\int \frac{4 {y}^{2} \cdot \mathrm{dy}}{{y}^{2} + 1} ^ 2$

After using $y = \tan z$ and $\mathrm{dy} = {\left(\sec z\right)}^{2} \cdot \mathrm{dz}$ transormation, it became

$\int \frac{4 {\left(\tan z\right)}^{2} \cdot {\left(\sec z\right)}^{2} \cdot \mathrm{dz}}{\sec z} ^ 4$

=$\int 4 {\left(\tan z\right)}^{2} \cdot {\left(\cos z\right)}^{2} \cdot \mathrm{dz}$

=$\int 4 {\left(\sin z\right)}^{2} \cdot \mathrm{dz}$

=$\int \left(2 - 2 \cos 2 z\right) \cdot \mathrm{dz}$

=$2 z - \sin 2 z + C$

=$2 z - \frac{2 \tan z}{{\left(\tan z\right)}^{2} + 1} + C$

After using $y = \tan z$ and $z = \arctan y$ inverse transforms, I found

$2 \arctan y - \frac{2 y}{{y}^{2} + 1} + C$

=$2 \arctan \sqrt{\frac{x}{2 - x}} - 2 \frac{\sqrt{\frac{x}{2 - x}}}{\frac{x}{2 - x} + 1} + C$

=$2 \arctan \sqrt{\frac{x}{2 - x}} - \sqrt{x \cdot \left(2 - x\right)} + C$

=$2 \arctan \sqrt{\frac{x}{2 - x}} - \sqrt{2 x - {x}^{2}} + C$