int{((logx-1))/(1+(logx)^2)}^2dx এর মান কত? 

Explanation:

Another Explanation (5): সমাধান: ধরি, \(I = \int \left(\frac{\log x - 1}{1 + (\log x)^2}\right)^2 dx\) এখন, \(\log x = t\) ধরি। সুতরাং, \(x = e^t\) এবং \(dx = e^t dt\). তাহলে, \(I = \int \left(\frac{t - 1}{1 + t^2}\right)^2 e^t dt = \int e^t \frac{(t-1)^2}{(1+t^2)^2} dt\) \( = \int e^t \frac{t^2 - 2t + 1}{(1+t^2)^2} dt = \int e^t \frac{t^2 + 1 - 2t}{(1+t^2)^2} dt\) \( = \int e^t \left(\frac{1}{1+t^2} - \frac{2t}{(1+t^2)^2}\right) dt\) এখন, আমরা জানি, \(\int e^x [f(x) + f'(x)] dx = e^x f(x) + C\) এখানে, \(f(t) = \frac{1}{1+t^2}\) সুতরাং, \(f'(t) = \frac{-2t}{(1+t^2)^2}\) তাহলে, \(I = \int e^t \left(\frac{1}{1+t^2} + \frac{-2t}{(1+t^2)^2}\right) dt = e^t \frac{1}{1+t^2} + C\) এখন, \(t = \log x\) বসিয়ে পাই, \(I = e^{\log x} \frac{1}{1+(\log x)^2} + C = x \frac{1}{1+(\log x)^2} + C\) \( = \frac{x}{1+(\log x)^2} + C\) সুতরাং, \(\int \left(\frac{\log x - 1}{1 + (\log x)^2}\right)^2 dx = \frac{x}{1+(\log x)^2} + C\) অতএব, নির্ণেয় মান \(\frac{x}{(\log x)^2 + 1} + c\) ✅