Evaluate $\lim\limits_{n\to+\infty}\int_{0}^{1}\ln(1-x^2+2x^{n}+x^{2n}){\rm d}x$. Problem
Evaluate $$\lim_{n\to+\infty}\int_{0}^{1}\ln(1-x^2+2x^{n}+x^{2n}){\rm d}x.$$
I guess it needs to apply Integral Mean Value Theorem or Integral Inequality . But I fail to find the upper-bound function and the lower-bound function. Who can offer a hint?Thanks!
 A: The integrals at hand can be rewritten as
$$\int_0^1 \log(1 - x^2 + 2x^n + x^{2n}) dx
= 2\int_0^1 \log(1+x^n) dx - \int_0^1 f_n(x) dx$$
where $f_n(x) = -\log\left(1 - \left(\frac{x}{1+x^n}\right)^2 \right)$. 
For the first piece, we can bound it as
$$0 \le \int_0^1 \log(1+x^n) dx \le \int_0^1 x^n dx = \frac{1}{n+1}$$
By squeezing, this leads to $\lim_{n\to\infty} \int_0^1 \log(1+x^n) dx = 0$.
For the second piece, notice for any fixed $x \in (0,1)$, we have
$$0 \le f_n(x) \le f_{n+1}(x) \le \infty
\quad\text{ and }\quad
\lim_{n\to\infty} f_n(x) = -\log(1-x^2)$$
Using Monotone converge theorem, we find
$$\lim_{n\to\infty} \int_0^1 f_n(x) dx = \int_0^1 \lim_{n\to\infty} f_n(x) dx
= -\int_0^1\log(1-x^2) dx = 2-\log 4$$
Combine these, we can conclude the integrals at hand converge and
$$\lim_{n\to\infty}\int_0^1\log(1-x^2 + 2x^n + x^{2n}) dx = 2\cdot 0 - (2-\log 4) = \log 4 - 2$$
A: An addendum to Achille's answer on how to evaluate the integral once the convergence difficulties have been addressed. In the region $|x| < 1$, \begin{align*} \int \log(1-x^2)\, dx &= x \log (1-x^2) + \int \frac{2x^2}{1-x^2}\, dx \tag{integration by parts} \\
&= x \log (1-x^2) + \int \left(\frac{1}{1+x} + \frac{1}{1-x} - 2\, \right) \tag{partial fractions}dx \\
&= (x+1) \log (1+x) + (x+1) \log (1-x) - 2x + C.
\end{align*}
Taking $0$ and $1$ as our limits of integration, we note that the last line equals $C$ at $x = 0$. At $x = 1$, the term $(x-1) \log (1-x)$ vanishes as $\lim_{x \to 0} x \log x$, and the remaining terms give $2 \log 2 - 2 + C$, so we have $$\int_0^1 \log(1-x^2)\, dx = \log 4 - 2.$$
