int_1^(e^2) dx/(x(1+logx)^2 এর মান কোনটি?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(1 + \log x = z\) তাহলে, \(\frac{1}{x} dx = dz\) যখন \(x = 1\), তখন \(z = 1 + \log 1 = 1 + 0 = 1\) যখন \(x = e^2\), তখন \(z = 1 + \log e^2 = 1 + 2 = 3\) সুতরাং, \(\int_1^{e^2} \frac{dx}{x(1+\log x)^2} = \int_1^3 \frac{dz}{z^2}\) \(= \int_1^3 z^{-2} dz\) \(= \left[ \frac{z^{-1}}{-1} \right]_1^3\) \(= \left[ -\frac{1}{z} \right]_1^3\) \(= -\frac{1}{3} - (- \frac{1}{1})\) \(= -\frac{1}{3} + 1\) \(= \frac{-1+3}{3}\) \(= \frac{2}{3}\) সুতরাং, \(\int_1^{e^2} \frac{dx}{x(1+\log x)^2} = \frac{2}{3}\)। 🎉