How to find limit $\lim_{n\to \infty}\left[(n+1)\int_{0}^{1}x^n\ln(x+1)dx\right]$ How to find this limit;
$$\lim_{n\to \infty}\left[(n+1)\int_{0}^{1}x^n\ln(x+1)dx\right]$$
What techniques can I use here? Thanks for reading and help.
This question was asked in GATE 2008.
 A: Hint:
You can use IBP to evaluate the integral as: $$\bigg[\frac{x^{n+1}\ln(x+1)}{n+1}\bigg]^1_0-\frac{1}{n+1}\int_0^1{\frac{x^{n+1}}{x+1}dx}$$
$$=\frac{1}{n+1}\bigg(\ln 2 -\int_0^1{\frac{x^{n+1}}{x+1}dx}\bigg)$$
This means you are looking for $$\lim_{n\to\infty}\bigg(\ln(2)-\int_0^1{\frac{x^{n+1}}{x+1}dx}\bigg)$$
Can you prove this integral goes to $0$?
A: Too long for a comment but added for your curiosity (even if too complex).
You have received good answers already so I shall focus on the asymptotics of
$$I_n=(n+1)\int_{0}^{1}x^n\ln(x+1)\,dx$$
First of all, if you know about the Gaussian hypergeometric function, there is an antiderivative
$$\int x^n\ln(x+1)\,dx=\frac{x^{n+1} (\, _2F_1(1,n+1;n+2;-x)+(n+1) \log (x+1)-1)}{(n+1)^2}$$ and, assuming $n>0$, using the bounds
$$\int_0^1 x^n\ln(x+1)\,dx=\frac{(n+1)\left( H_{\frac{n}{2}}- H_{\frac{n-1}{2}}\right)+n \log (4)+2\log (2)-2}{2
   (n+1)^2}$$ Using the asymptotics of harmonic numbers
$$H_p=\gamma +\log(p) +\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^4}\right)$$ and continuing with Taylor series, we should end with
$$I_n=\log (2)-\frac{1}{2 n}+\frac{3}{4
   n^2}+O\left(\frac{1}{n^3}\right)$$
For sure, the same result would be obtained using Rhys Hughes's answer since 
$$\int \frac{x^{n+1}}{x+1}\,dx=-\frac{x^{n+1} (\, _2F_1(1,n+1;n+2;-x)-1)}{n+1}$$
$$\int_0^1 \frac{x^{n+1}}{x+1}\,dx=\frac{1}{2} \left(H_{\frac{n+1}{2}}-H_{\frac{n}{2}}\right)$$
