int_0^1(ln(x+1)/(x+1))dx=?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(I = \int_{0}^{1} \frac{\ln(x+1)}{x+1} dx\) এখন, \(x+1 = t\) ধরি। সুতরাং, \(dx = dt\) হবে। যখন \(x = 0\), তখন \(t = 1\) এবং যখন \(x = 1\), তখন \(t = 2\) হবে। তাহলে, \(I = \int_{1}^{2} \frac{\ln(t)}{t} dt\) এখন, \(\ln(t) = u\) ধরি। সুতরাং, \(\frac{1}{t} dt = du\) হবে। যখন \(t = 1\), তখন \(u = \ln(1) = 0\) এবং যখন \(t = 2\), তখন \(u = \ln(2)\) হবে। তাহলে, \(I = \int_{0}^{\ln(2)} u du\) \(I = \left[ \frac{u^2}{2} \right]_{0}^{\ln(2)}\) \(I = \frac{(\ln(2))^2}{2} - \frac{0^2}{2}\) \(I = \frac{1}{2} (\ln 2)^2\) 🎉 অতএব, \(\int_{0}^{1} \frac{\ln(x+1)}{x+1} dx = \frac{1}{2} (\ln 2)^2\) 🥳