For large $n$, show that $\int\limits_{0}^{1}\frac{nx^{n-1}dx}{1+x^2} $ nearly equals $\frac{1}{2}$.

Question

For large $n$, show that $$\int\limits_{0}^{1}\frac{nx^{n-1}dx}{1+x^2} $$ nearly equals $\frac{1}{2}$.

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Integrating by parts we get $$\int\limits_{0}^{1}\frac{nx^{n-1}dx}{1+x^2}=\Bigg(\frac{x^n}{1+x^2}\Bigg)^{1}_{0} - \int\limits_{0}^1 \frac{2x^{n+1}}{(1+x^2)^2}dx$$ The integral $$\int\limits_{0}^1 \frac{2x^{n+1}}{(1+x^2)^2}dx\leq\int\limits_{0}^1 {2x^{n+1}}dx =\frac{2}{n+2}$$ which $\rightarrow 0$ as $n\rightarrow \infty$
Hence the value of the main integral is $1$. Where am I wrong ?

Answer 1

As commented:

$$\left.\left(\frac{x^n}{1+x^2}\right)\right|_0^1=\frac1{1+1}-\frac0{1+0}=\frac12$$