Find the value of : $\lim_{n\to\infty}[(n+1)\int_{0}^{1}x^{n}\ln(1+x)dx]$.

As you've shown, using integration by parts, one has $$\lim_{n\to\infty}\int_0^1x^n\ln(1+x)\:dx=\ln(2)-\lim_{n\to\infty}\int_0^1\frac{x^{n+1}}{1+x}\:dx.$$ Hence, one we calculate the limit of the integral, we'll have the solution. Note that $$\left|\int_0^1\frac{x^{n+1}}{1+x}\:dx\right|\leq\int_0^1\left|x^{n+1}\right|\:dx.$$ We have this bound since $1+x\geq1$, so $\frac{1}{1+x}\leq1$. Now it is easy to see that the integral on the right-hand side has a limit of zero, and therefore $$\lim_{n\to\infty}\int_0^1x^n\ln(1+x)\:dx=\ln(2).$$

Hint: Now you need to prove the second term $\int_0^1 \frac{x^{n+1}}{1+x}dx$ has limit zero when $n$ goes to infinity.

Try to bound it from above with another simpler integral that goes to zero when $n\to \infty$.