# Quick Arc Length Question

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!

• 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. – Lee Mosher Jan 19 at 15:31
• And if this makes your question not so "quick", that's fine. – Lee Mosher Jan 19 at 15:33
• Edits should've been made. – joe Jan 19 at 15:36
• What makes you think that this is "quick" ? – Yves Daoust Jan 19 at 15:38

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 $$\approx 3.3197.$$
• @Kevin Mathematica's NIntegrate. Read the documentation – K.defaoite Jan 19 at 17:03