$\lim\limits_{n\to\infty}\int_0^2 \frac{nx^{n-1}e^{-x^n}}{1+x}dx=1-\lim\limits_{n\to\infty}\int_0^2\frac{e^{-x^n}}{(1+x)^2}dx$

Question

How can I show $\lim\limits_{n\to\infty}\int_0^2 \frac{nx^{n-1}e^{-x^n}}{1+x}dx=1-\lim\limits_{n\to\infty}\int_0^2\frac{e^{-x^n}}{(1+x)^2}dx$?
It is $\frac{d}{dx}x^n=nx^{n-1}$ so I tried to solve it with substitution but this idea did not lead to the goal.

Answer 1

Observe that: $$ \frac{\mathrm{d}}{\mathrm{d}x} e^{-x^n} = -nx^{n-1}e^{-x^n} $$ so you can integrate by parts to get: \begin{align*} \int_0^2 \frac{nx^{n-1}e^{-x^n}}{1 + x} \, \mathrm{d}x &= \left[-\frac{e^{-x^n}}{1 + x}\right]_0^2 - \int_0^2 -\frac{e^{-x^n}}{-(1 + x)^2} \, \mathrm{d}x \\ &= e^{-2^n} - \int_0^2 \frac{e^{-x^n}}{(1 + x)^2} \, \mathrm{d}x \end{align*} Letting $n \to \infty$, we have that $\lim_{n \to \infty} e^{-2^n} = 1$.