inte^x{1/(1-x)+1/(1-x)^2}dx= কত?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(I = \int e^x \left(\frac{1}{1-x} + \frac{1}{(1-x)^2}\right) dx\) এখানে, \(f(x) = \frac{1}{1-x}\) এবং \(f'(x) = \frac{1}{(1-x)^2}\) আমরা জানি, \(\int e^x (f(x) + f'(x)) dx = e^x f(x) + c\) সুতরাং, \(I = e^x \cdot \frac{1}{1-x} + c = \frac{e^x}{1-x} + c\) অতএব, \(\int e^x \left(\frac{1}{1-x} + \frac{1}{(1-x)^2}\right) dx = \frac{e^x}{1-x} + c\) 🎉🎉