Question

√1-√x/1+√x, x ∈ (0,1),Integrate it with respect to x.

Answer 1

please see the attachment

Answer 2

Answer:

2 - $\frac{\pi}{2}$

Step-by-step explanation:

In this question

We have been integrate this equation

⇒ $\int_{0}^{1} [\sqrt\frac{1 - (\rt x)}{1 - (\sqrt x)}]\times dx$

⇒ $\int_{0}^{1} [\sqrt\frac{1 - (\sqrt x)}{1 + \sqrt x)}\times \frac{1 - (\sqrt x)}{1 - (\sqrt x)}]\times dx$

⇒ $x\int_{0}^{1} [\frac{1 - (\sqrt x)}{\sqrt (1 - x)}]\times dx$

⇒ $x\int_{0}^{1} [\frac{1}{\sqrt (1 - x)}]\times dx - x\int_{0}^{1} [\frac{(\sqrt x)}{\sqrt (1 - (x)}]\times dx$

⇒ $\textrm[-2\sqrt(1 - x)]_{0}^{1} - x\int_{0}^{1} [\frac{(\sqrt x)}{\sqrt (1 - (x)}]\times dx$

Put x = $sin^{2}\Theta$

     dx = $2\times sin\Theta\times cos\Theta\times d\Theta$

⇒ 2 - $x\int_{0}^{\frac{\pi}{2}} [\frac{sin\Theta}{cos\Theta}\times 2\times sin\Theta\times cos\Theta]\times d\Theta$

⇒ 2 - $x\int_{0}^{\frac{\pi}{2}}2[sin^{2}\Theta]\times d\Theta$

⇒ 2 - $x\int_{0}^{\frac{\pi}{2}}2[1 - cos2\Theta]\times d\Theta$

⇒ 2 - $\textrm[\Theta -\frac{sin2\Theta}{2}]_{0}^{\frac{\pi}{2}}$

⇒ 2 - $\frac{\pi}{2}$