# Limit of $(\int\limits_0^n (1+\arctan^2x )\,dx )^ {\frac{1}{n}}$

I need to find out this limit. Could someone help me? $$\lim_{n\rightarrow \infty} \Big(\int\limits_0^n (1+\arctan^2x )\,dx \Big)^ {\frac{1}{n}}$$ = ?

I have tried taking logarithm then calculating the integral of $$arctan^{2}(x)$$, it got worse, it seems to me that there is some shorter solution..

• What about your thoughts on the problem? Did you try to take logarithm of both sides, and use l'Hopital rule? Please include it in your question – Jakobian Oct 10 '18 at 18:17
• Do you know the limit of $n^{1/n}$ as $n\to+\infty$? – mickep Oct 10 '18 at 18:27
• @Jakobian, have tried taking logarithm then calculating the integral of $arctan^{2}(x)$, it got worse, it seems to me that there is some shorter solution.. – Emathke Oct 10 '18 at 18:31
• @Emathke I actually said "Please include it in your question" to avoid you writing it to me in the comments – Jakobian Oct 10 '18 at 18:34

Your integrand is bounded as $$1\leq 1+\arctan^2x\leq 1+\frac{\pi^2}{4}.$$ Thus, your integral is between $$n$$ and $$n(1+\pi^2/4)$$. Taking the $$n$$th root and using the squeeze theorem for limits together with the facts that $$\lim_{n\to+\infty}n^{1/n}=1 \qquad\text{and}\qquad \lim_{n\to+\infty}a^{1/n}=1$$ should get you to the goal.