How do I know if this limit even converges? So, there is this question which I found quite new according to my experience, and it is as follows.

Assuming that interchange of limit and integration is permissible, evaluate the following: $$\lim_{n \to \infty} {\int_0^1 \frac{nx^{n -1}}{1 + x}dx}, \quad 0 < x < 1$$

Now, I first tried integration by parts, and it doesn't give me more than an expression which is also nearly impossible to evaluate (maybe I'm doing it wrong, though) -

$$\lim_{n \to \infty} \left|\frac{1}{1+x}\int nx^{n - 1}dx  +  \int\frac{x^n}{(1 + x)^2}dx\right|_0^1$$

I'm not sure how to proceed from there, so I tried this instead:

$$\lim_{n \to \infty} {\int_0^1 {nx^{n - 1}(1 + x)^{-1}}dx}$$

Expanding $(1 + x)^{-1}$ using Taylor Series:

$$\lim_{n \to \infty} {\int_0^1 {nx^{n - 1}(1 - x + x^2 - x^3 + x^4 - x^5 \ldots)}dx}$$
$$ = \lim_{n \to \infty} {\int_0^1 {(nx^{n - 1} - nx^n + nx^{n + 1} - nx^{n + 2} + nx^{n + 3} - n^{x + 4} \ldots)}dx}$$
$$\lim_{n \to \infty} \left(1 - \frac{n}{n+1} + \frac{n}{n+2} - \frac{n}{n +3} + \frac{n}{n +4} - \frac{n}{n + 5}\ldots\right)$$

I think that diverges. Then, I tried putting the expression in Desmos and increased the $n$ value slowly upto $500$, $1000$ and similar big numbers - which all returned values around $0.5$.
The question was an multiple-choice based and there were both of them listed as options.
I'm not sure who is correct, the calculator, or me.
It would be a huge help if someone looks into this. Thank you in advance.
 A: OK. Use integration by parts. Rewrite $I_n = \displaystyle \int_{0}^1 \dfrac{d(x^n)}{1+x}= \dfrac{x^n}{1+x}|_{x=0}^1+ \displaystyle \int_{0}^1\dfrac{x^n}{(1+x)^2}dx= \dfrac{1}{2}+J_n$, and $0 \le J_n \le \displaystyle \int_{0}^1 x^ndx= \dfrac{1}{n+1}$. Hence $J_n \to 0$, and $I_n \to \dfrac{1}{2}$. So if you got $0.5$ then  you are right.
A: For your curiosity.
$$I_n=\int n\frac{x^{n -1}}{1 + x}dx=n \left(\frac{x^n}{n}-\frac{x^{n+1}}{n+1} \, _2F_1(1,n+1;n+2;-x)\right)$$ where appears the gaussian hypergeometric function.
Integrating between the bounds, this gives the simple
$$J_n=\int_0^1 n\frac{x^{n -1}}{1 + x}dx=1+\frac{1}{2} n \left(H_{\frac{n-1}{2}}-H_{\frac{n}{2}}\right)$$ Using the asymptotics of generalized harmonic numbers
$$J_n=\frac{1}{2}+\frac{1}{4 n}-\frac{1}{8 n^3}+O\left(\frac{1}{n^5}\right)$$
A: Here's another method in the same spirit as a couple comments.  We trade off the use of integration by parts in favor of just using monotonicity of the integral.
Fix any $y$ such that $0 < y < 1$ and split the integral into two as
$$\int_0^1 \frac{nx^{n-1}}{1+x}\,dx = \int_0^y \frac{nx^{n-1}}{1+x}\,dx + \int_y^1\frac{nx^{n-1}}{1+x}\,dx$$
Then estimate each integral independently:

*

*On $0 \leq x \leq y < 1$, we have $0 \leq \frac{nx^{n-1}}{1+x} \leq nx^{n-1}$, so $$0 \leq \int_0^y \frac{nx^{n-1}}{1+x}\,dx \leq \int_0^y nx^{n-1}\,dx = y^n$$ and since $y^n \to 0$ we have $$\lim_{n\to\infty} \int_0^y \frac{nx^{n-1}}{1+x}\,dx = 0$$

*On $0 < y \leq x \leq 1$, we have $\frac{nx^{n-1}}{2} \leq \frac{nx^{n-1}}{1+x} \leq \frac{nx^{n-1}}{1+y}$, so $$\frac{1}{2}(1 - y^n) = \int_y^1\frac{nx^{n-1}}{2}\,dx \leq \int_y^1\frac{nx^{n-1}}{1+x}\,dx \leq \int_y^1\frac{nx^{n-1}}{1+y}\,dx = \frac{1 - y^n}{1+y}.$$  As once again, $y^n \to 0$, we have $$\frac{1}{2} \leq \liminf_{n\to\infty} \int_y^1 \frac{nx^{n-1}}{1+x}\,dx \leq \limsup_{n\to\infty} \int_y^1 \frac{nx^{n-1}}{1+x}\,dx \leq \frac{1}{1+y}$$

Putting it back in terms of the original integral, we've shown $$\frac{1}{2} \leq \liminf_{n\to\infty} \int_0^1 \frac{nx^{n-1}}{1+x}\,dx \leq \limsup_{n\to\infty} \int_0^1 \frac{nx^{n-1}}{1+x}\,dx \leq \frac{1}{1+y}$$
which now only involves $0 < y < 1$ on the upper bound, and letting $y \to 1^{-}$ shows $$\lim_{n\to\infty} \int_0^1 \frac{nx^{n-1}}{1+x}\,dx = \frac{1}{2}$$

The other reason I mention this method is that it generalizes nicely.  We've used monotonicity of $\frac{1}{1+x}$ to keep everything clean, but even this is actually unnecessary.  The same argument may be made with any function $f$ which is continuous on $[0,1]$ (or even just $f \in L^{\infty}([0,1])$ which is left-continuous at $1$) to give $\int_0^1 nx^{n-1}f(x)\,dx \to f(1)$.
A: Substitute $x=t^{1/n}$ to obtain
$$
\int_0^1 {\frac{{nx^{n - 1} }}{{1 + x}}dx}  = \int_0^1 {\frac{{dt}}{{1 + t^{1/n} }}} 
$$
Now, since
$$
0 \le \frac{1}{{1 + t^{1/n} }} \le 1,
$$
we can apply the dominated convergence theorem to deduce
$$
\mathop {\lim }\limits_{n \to  + \infty } \int_0^1 {\frac{{nx^{n - 1} }}{{1 + x}}dx}  = \mathop {\lim }\limits_{n \to  + \infty } \int_0^1 {\frac{{dt}}{{1 + t^{1/n} }}}  = \int_0^1 {\mathop {\lim }\limits_{n \to  + \infty } \frac{{dt}}{{1 + t^{1/n} }}}  = \int_0^1 {\frac{{dt}}{2}}  = \frac{1}{2}.
$$
