int_{0}^{1} sqrt{frac{1+x}{1-x}} , dx = | vedclass

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(A) माना $ I = \int_{0}^{1} \sqrt{\frac{1+x}{1-x}} \, dx $. समाकल्य का परिमेयकरण करने पर: $ I = \int_{0}^{1} \sqrt{\frac{(1+x)^2}{(1-x)(1+x)}} \, dx =

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$ \frac{\pi}{2} + 1 $

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(A) माना $ I = \int_{0}^{1} \sqrt{\frac{1+x}{1-x}} \, dx $. समाकल्य का परिमेयकरण करने पर: $ I = \int_{0}^{1} \sqrt{\frac{(1+x)^2}{(1-x)(1+x)}} \, dx = \int_{0}^{1} \frac{1+x}{\sqrt{1-x^2}} \, dx $. समाकलन को अलग करने पर: $ I = \int_{0}^{1} \frac{1}{\sqrt{1-x^2}} \, dx + \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} \, dx $. प्रथम भाग का मूल्यांकन: $ \int_{0}^{1} \frac{1}{\sqrt{1-x^2}} \, dx = [\sin^{-1} x]_{0}^{1} = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2} $. द्वितीय भाग का मूल्यांकन: माना $ 1-x^2 = t $,तब $ -2x \, dx = dt $,अर्थात $ x \, dx = -\frac{1}{2} \, dt $. जब $ x=0, t=1 $ और जब $ x=1, t=0 $. $ \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} \, dx = \int_{1}^{0} \frac{-1/2}{\sqrt{t}} \, dt = \frac{1}{2} \int_{0}^{1} t^{-1/2} \, dt = \frac{1}{2} [2\sqrt{t}]_{0}^{1} = \frac{1}{2} (2) = 1 $. अतः,$ I = \frac{\pi}{2} + 1 $.

$ \int_{0}^{1} \sqrt{\frac{1+x}{1-x}} \, dx = $

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