intsqrt((1+x)/(1-x)) dx = কত?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(I = \int \sqrt{\frac{1+x}{1-x}} dx\) \(\sqrt{\frac{1+x}{1-x}}\) কে সরল করি: \( \sqrt{\frac{1+x}{1-x}} = \sqrt{\frac{(1+x)(1+x)}{(1-x)(1+x)}} = \sqrt{\frac{(1+x)^2}{1-x^2}} = \frac{1+x}{\sqrt{1-x^2}} \) তাহলে, \(I = \int \frac{1+x}{\sqrt{1-x^2}} dx = \int \frac{1}{\sqrt{1-x^2}} dx + \int \frac{x}{\sqrt{1-x^2}} dx \) এখানে, \(I = I_1 + I_2\) \(I_1 = \int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1}x + c_1\) 🥳 এখন, \(I_2 = \int \frac{x}{\sqrt{1-x^2}} dx\) ধরি, \(1-x^2 = u\) তাহলে, \(-2x dx = du\) বা, \(x dx = -\frac{1}{2} du\) সুতরাং, \(I_2 = \int \frac{-\frac{1}{2}}{\sqrt{u}} du = -\frac{1}{2} \int u^{-\frac{1}{2}} du = -\frac{1}{2} \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + c_2 = - \sqrt{u} + c_2 = - \sqrt{1-x^2} + c_2\) 😎 অতএব, \(I = I_1 + I_2 = \sin^{-1}x - \sqrt{1-x^2} + c\) , যেখানে \(c = c_1 + c_2\) একটি ধ্রুবক। সুতরাং, \(\int \sqrt{\frac{1+x}{1-x}} dx = \sin^{-1}x - \sqrt{1-x^2} + c\) 🎉