sqrt((1+x)/(1-x)) এর ইন্টিগ্রেশন -

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

এখন, \(\sqrt{\frac{1+x}{1-x}} = \sqrt{\frac{(1+x)(1+x)}{(1-x)(1+x)}} = \sqrt{\frac{(1+x)^2}{1-x^2}} = \frac{1+x}{\sqrt{1-x^2}}\)

সুতরাং, \(I = \int \frac{1+x}{\sqrt{1-x^2}} \, dx = \int \frac{1}{\sqrt{1-x^2}} \, dx + \int \frac{x}{\sqrt{1-x^2}} \, dx\)

আমরা জানি, \(\int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1}x + c_1\)

এখন, \(I_2 = \int \frac{x}{\sqrt{1-x^2}} \, dx\) বিবেচনা করি।

ধরি, \(1-x^2 = u\), তাহলে \(-2x \, dx = du\), সুতরাং \(x \, dx = -\frac{1}{2} du\)

অতএব, \(I_2 = \int \frac{-\frac{1}{2}}{\sqrt{u}} \, du = -\frac{1}{2} \int u^{-\frac{1}{2}} \, du = -\frac{1}{2} \cdot \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + c_2 = -u^{\frac{1}{2}} + c_2 = -\sqrt{1-x^2} + c_2\)

তাহলে, \(I = \sin^{-1}x - \sqrt{1-x^2} + c\), যেখানে \(c = c_1 + c_2\)

সুতরাং, \(\int \sqrt{\frac{1+x}{1-x}} \, dx = \sin^{-1}x - \sqrt{1-x^2} + c\) 🥳🥳🥳