যোগজ নির্ণয় কর  :  int(dx)/(xsqrt(x^2+1)) 

Explanation:

Another Explanation (5): যোগজ নির্ণয়: \(\int \frac{dx}{x\sqrt{x^2+1}}\) আমরা \(x = \frac{1}{t}\) প্রতিস্থাপন করি। তাহলে, \(dx = -\frac{1}{t^2} dt\) সুতরাং, \(\int \frac{dx}{x\sqrt{x^2+1}} = \int \frac{-\frac{1}{t^2} dt}{\frac{1}{t}\sqrt{\frac{1}{t^2}+1}} \) \( = \int \frac{-\frac{1}{t^2} dt}{\frac{1}{t}\sqrt{\frac{1+t^2}{t^2}}} \) \( = \int \frac{-\frac{1}{t^2} dt}{\frac{1}{t} \frac{\sqrt{1+t^2}}{t}} \) \( = \int \frac{-\frac{1}{t^2} dt}{\frac{\sqrt{1+t^2}}{t^2}} \) \( = \int \frac{-1}{\sqrt{1+t^2}} dt \) \( = -\int \frac{1}{\sqrt{1+t^2}} dt \) আমরা জানি, \(\int \frac{1}{\sqrt{x^2+1}} dx = \sinh^{-1}(x) + C\) অথবা \(\ln(x+\sqrt{x^2+1})+C\) সুতরাং, \( = -\sinh^{-1}(t) + C \) \( = -\ln(t+\sqrt{t^2+1}) + C \) যেহেতু \(x = \frac{1}{t}\), তাই \(t = \frac{1}{x}\). অতএব, \( = -\ln(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1}) + C \) \( = -\ln(\frac{1}{x}+\frac{\sqrt{1+x^2}}{x}) + C \) \( = -\ln(\frac{1+\sqrt{1+x^2}}{x}) + C \) \( = -\ln(1+\sqrt{x^2+1}) + \ln(x) + C \) \( = \ln(x) - \ln(1+\sqrt{x^2+1}) + C \) \( = \ln(\frac{x}{1+\sqrt{x^2+1}}) + C \) 🥳 আমরা আরও সরল করতে পারি: \( \frac{x}{1+\sqrt{x^2+1}} = \frac{x(1-\sqrt{x^2+1})}{(1+\sqrt{x^2+1})(1-\sqrt{x^2+1})} \) \( = \frac{x(1-\sqrt{x^2+1})}{1-(x^2+1)} \) \( = \frac{x(1-\sqrt{x^2+1})}{-x^2} \) \( = \frac{\sqrt{x^2+1}-1}{x} \) সুতরাং, \( = \ln(\frac{\sqrt{x^2+1}-1}{x}) + C \) 🤩 \( = -\ln(\frac{x}{\sqrt{x^2+1}-1}) + C \) সুতরাং নির্ণেয় যোগজ হল: \(\ln\left(\frac{x}{1+\sqrt{x^2+1}}\right) + C\) অথবা \(-\ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1}\right) + C\) অথবা \(\ln\left(\frac{\sqrt{x^2+1}-1}{x}\right) + C\) 😎