Given a limit: $$\lim_{n\to+\infty}\frac{1}{n+1}\int_{0}^{n}\arctan(x)\,dx = \alpha$$ Find the value of $\alpha$.
Well, the inner integral equals: $$\int_{0}^{n}\arctan(x)\,dx = n\cdot\arctan(n)\bigr|_{0}^{+\infty} - \frac{1}{2}\ln\left(n^2+1\right)\Bigr|_{0}^{+\infty}$$ Then, I rewrite the limit: $$\lim_{n\to+\infty}\frac{n\cdot\arctan(n)\bigr|_{0}^{+\infty} - \frac{1}{2}\ln\left(n^2+1\right)\bigr|_{0}^{+\infty}}{n+1}$$ By inserting $+\infty$ and $0$ in the limit I obtain $\alpha = 0$. However, this is not a correct answer.
How would I proceed? Thank you.
Hint. There is no need to evaluate the integral. By using L'Hopital and the Fundamental Theorem of calculus, we have that $$\lim_{t\to+\infty}\frac{1}{t+1}\int_{0}^{t}\arctan(x)dx= \lim_{t\to+\infty}\arctan(t).$$