${\displaystyle {\int }_0^1\left(\frac{\ln (x+1)}{x+1}\right)dx=?}$

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 \( u = \ln(1+1) = \ln(2) \)

অতএব, সমাকলনটি হবে:

\( \int_{0}^{\ln(2)} u \, du \)

\( = \left[ \frac{u^2}{2} \right]_{0}^{\ln(2)} \)

\( = \frac{(\ln(2))^2}{2} - \frac{0^2}{2} \)

\( = \frac{1}{2} (\ln(2))^2 \)

সুতরাং, \( \int_{0}^{1} \frac{\ln(x+1)}{x+1} dx = \frac{1}{2} (\ln 2)^2 \)

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