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AI-তৈরি নমুনা উত্তর

ধরি, $I = \int \frac{dx}{\sqrt{x} + \sqrt{1-x}}$

হর ও লবকে $(\sqrt{x} - \sqrt{1-x})$ দ্বারা গুণ করে পাই,
$I = \int \frac{\sqrt{x} - \sqrt{1-x}}{(\sqrt{x} + \sqrt{1-x})(\sqrt{x} - \sqrt{1-x})} dx = \int \frac{\sqrt{x} - \sqrt{1-x}}{2x - 1} dx$

ধরি, $x = \sin^2\theta$
তাহলে, $dx = 2\sin\theta\cos\theta\, d\theta = \sin 2\theta\, d\theta$
এবং $2x - 1 = 2\sin^2\theta - 1 = -\cos 2\theta$

সুতরাং,
$I = \int \frac{\sin\theta - \cos\theta}{-\cos 2\theta} (2\sin\theta\cos\theta)\, d\theta$
$= \int \frac{\cos\theta - \sin\theta}{\cos^2\theta - \sin^2\theta} (2\sin\theta\cos\theta)\, d\theta$
$= \int \frac{2\sin\theta\cos\theta}{\cos\theta + \sin\theta} d\theta$
$= \int \frac{(\cos\theta + \sin\theta)^2 - 1}{\cos\theta + \sin\theta} d\theta$
$= \int (\cos\theta + \sin\theta) d\theta - \int \frac{d\theta}{\sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right)}$
$= \sin\theta - \cos\theta - \frac{1}{\sqrt{2}} \int \sec\left(\theta - \frac{\pi}{4}\right) d\theta$
$= \sin\theta - \cos\theta - \frac{1}{\sqrt{2}} \ln \left| \sec\left(\theta - \frac{\pi}{4}\right) + \tan\left(\theta - \frac{\pi}{4}\right) \right| + C$

যেহেতু $\sin\theta = \sqrt{x}$ এবং $\cos\theta = \sqrt{1-x}$:
$\tan\left(\theta - \frac{\pi}{4}\right) = \frac{\tan\theta - 1}{\tan\theta + 1} = \frac{\sqrt{x} - \sqrt{1-x}}{\sqrt{x} + \sqrt{1-x}}$
এবং $\sec\left(\theta - \frac{\pi}{4}\right) = \frac{\sqrt{2}}{\cos\theta + \sin\theta} = \frac{\sqrt{2}}{\sqrt{x} + \sqrt{1-x}}$

অতএব,
$I = \sqrt{x} - \sqrt{1-x} - \frac{1}{\sqrt{2}} \ln \left| \frac{\sqrt{2} + \sqrt{x} - \sqrt{1-x}}{\sqrt{x} + \sqrt{1-x}} \right| + C$