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

Explanation:

Another Explanation (5): সমাধান: ধরি, \( u = \ln(x+1) \) তাহলে, \( du = \frac{1}{x+1} dx \) যখন \( x = 0 \), তখন \( u = \ln(0+1) = \ln(1) = 0 \) যখন \( x = 1 \), তখন \( u = \ln(1+1) = \ln(2) \) সুতরাং, ইন্টিগ্রালটি হবে: \( \int_0^{\ln(2)} u \, du \) \( = \left[ \frac{u^2}{2} \right]_0^{\ln(2)} \) \( = \frac{(\ln(2))^2}{2} - \frac{0^2}{2} \) \( = \frac{1}{2} (\ln(2))^2 \) অতএব, \( \int_0^1 \frac{\ln(x+1)}{x+1} dx = \frac{1}{2} (\ln 2)^2 \) 🎉