Equivalent of the integral of a function sequence I gave an exercise to a student last wednesday, but I had some trouble finding a correct solution to the last question !
For $n\ge2$, let $f_n$ be the function
$$f_n:[0,+\infty[\to\mathbb R,\,t\mapsto \frac{t^n}{1+t+t^{n-1}}$$
The first question asks the punctual limit of the sequence $(f_n)$, no problem, it's the function 
$$f:x\mapsto\begin{cases}0&\text{if }0\le x<1 \\
                         \frac{1}{3}&\text{if }x=1 \\
                         t&\text{if }x>1 \end{cases}$$
The convergence cannot be uniform due too the fact that the limit is not continue.
Now by dominated convergence theorem (not usable at this time of the year), or a simple majoration, it's easy to prove that
$$\lim_{n\to\infty} \int_0^1 f_n=\int_0^1 f$$
Problem I had was to find an equivalent of $\int_0^1 (f_n-f)$. A usual technique is to make an integration by parts, but that leads to a troubling $n$ from the derivation of the denominator which doesn't permit any conclusion.
Another technique is to guess the equivalent by getting rid of troubling terms. I thought of $\int_0^1 \frac{t^n}{1+t}{\rm d}t$, or $\int_0^1 \frac{t^n}{3}{\rm d}t$, but I can't find a good majorant of the difference.
Do you have any thoughts to share about this ?
Thanks.

HINT:
The substitution $t\to t^{1/n}$ yields
$$\int_0^1 \frac{t^n}{1+t+t^{n-1}}\,dt=\frac1n\int_0^1 \frac{t^{1/n}}{1+t^{1/n}+t^{(n-1)/n}}\,dt\le \frac1n$$

Alternatively, simply note that $\left|\int_0^1 \frac{t^n}{1+t+t^{n-1}}\,dt\right|\le \int_0^1 t^n\,dt=\frac{1}{n+1}$.

EDIT:  The OP is requesting the first-order asymptotic term for the integral of interest
To develop the first term in the asymptotic (large $n$) expansion of the integral of interest we enforce the substitution $t=e^{-x/(n-1)}$.  Then, we have
$$n\int_0^1 \frac{t^n}{1+t+t^{n-1}}\,dt=\frac{n}{n-1}\int_0^\infty \frac{e^{-(n+1)x/(n-1)}}{1+e^{-x}+e^{-x/(n-1)}}\,dx \tag 1$$
As $n\to \infty$, the limit of the integral on the right-hand side of $(1)$ is given by
$$\lim_{n\to \infty}\int_0^\infty \frac{e^{-x}}{2+e^{-x}}\,dx=\log(3/2)$$
Therefore, we see that 
$$\lim_{n\to \infty}n\int_0^1 \frac{t^n}{1+t+t^{n-1}}\,dt=\log(3/2)$$
We can write therefore that
$$\int_0^1 \frac{t^n}{1+t+t^{n-1}}\,dt\sim \frac{\log(3/2)}{n}$$
