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So I was given the following prompt:

Let $f$ be the function satisfying $f(0)=0$ and $f'(x)=\frac{\ln(x+2)}{x^2+1}$ for $x>-2$. What is the length of the graph of $y=f(x)$ over the closed interval $[0,3]$?

I guess I'm a bit confused over what the length formula would look like in this example and how this formula would be evaluated. I understand that the length formula would look something like the following: $$L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx,$$ but I'm not seeing an easy integration when this formula is plugged into the equation. I ended up getting something like: $$L=\int_0^3\sqrt{1+\left[\frac{\ln(x+2)}{(x^2+1)}\right]^2}\,dx.$$ I'm not really understanding how I'd go about integrating this or what an answer would look like here. Any help/clarification would be appreciated!

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    $\begingroup$ You should show us what you get when "this formula is plugged into the equation", and explain how you tried to find an "easy integration"; otherwise we are left to make unfounded guesses as to what troubles you encountered. $\endgroup$ – Lee Mosher Jan 19 at 15:31
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    $\begingroup$ And if this makes your question not so "quick", that's fine. $\endgroup$ – Lee Mosher Jan 19 at 15:33
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    $\begingroup$ Edits should've been made. $\endgroup$ – joe Jan 19 at 15:36
  • $\begingroup$ What makes you think that this is "quick" ? $\endgroup$ – Yves Daoust Jan 19 at 15:38
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Basically you get the integral $$\int_0^3 \sqrt{1+\left(\frac{\ln(x+2)}{x^2+1}\right)^2}\mathrm{d}x$$ Numerical integration yields $\approx 3.3197.$

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    $\begingroup$ Can I ask what numerical technique you used? $\endgroup$ – Hyperkähler Jan 19 at 15:45
  • $\begingroup$ @Kevin Mathematica's NIntegrate. Read the documentation $\endgroup$ – K.defaoite Jan 19 at 17:03

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