Is my solution correct? If not, what is the right way to do this and where is my mistake? $$\lim_{x\to\infty}\frac{\int_0^x (\arctan^2 t )dt}{\sqrt{x^2+1}}$$ (L`Hopital.) $$\lim_{x\to\infty}\frac{F'(x)-F'(0)}{2x\cdot\frac{1}{2\sqrt{x^2+1}}}$$ $$\lim_{x\to\infty}\frac{\arctan^2(x)-\arctan^2(0)}{\frac{1}{\sqrt{1+\frac{1}{x^2}}}}$$ $$\frac{\pi^2}{4}$$ Thank you.
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Shold it not be $$\int_{0}^{x}(\arctan(t))^2dt$$? – Dr. Sonnhard Graubner May 22 '19 at 14:13
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@Dr.SonnhardGraubner Isn't it the same as $$\int_0^x(\arctan^2(t))dt$$? – fragileradius May 22 '19 at 14:15
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Your formula above isn't the same – Dr. Sonnhard Graubner May 22 '19 at 14:16
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It looks well in my opinion – joseabp91 May 22 '19 at 14:16
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ok, then see here https://math.stackexchange.com/questions/109105/limit-of-integration-cant-be-the-same-as-variable-of-integration – Dr. Sonnhard Graubner May 22 '19 at 14:18
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@Dr.SonnhardGraubner I understand now what you're talking about! I'll fix the post, thank you. – fragileradius May 22 '19 at 14:24
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3@fragileradius You need to justify that you can use L'Hôpital's rule, in particular that $\lim_{x \to \infty} \int_0^ x (\arctan t)^2 dt = \infty$. Also, you don't need subtract $F'(0)$... the derivative of $\int_a^x f(t) dt$ is equal to $f(x)$, regardless of $a$. – PierreCarre May 22 '19 at 14:32