<s>[INST] Arclength, finding definite integral Find the length of the segment between $x=5$ and $x=2$ for the following curve.
$$f(x)= \frac{1}{2}x^2-\frac{1}{4}\log\left(x\right)$$
My working so far.
$$f'\left(x\right)=x-\frac{1}{4x}.$$
Then
$$L = \int _2^5\:\sqrt{1+\left(f'\left(x\right)\right)^2}\,dx
=\int _2^5\:\sqrt{x^2+\frac{1}{2}+\frac{1}{16x^2}}\,dx.$$
However I have no idea how to find the solution to the definite integral.
 [/INST] $$L = \int _2^5\:\sqrt{1+\left(f'\left(x\right)^2\right)}
\\
\\=\int _2^5\:\sqrt{1+\left(x-\frac{1}{4x}\right)^2}\\
\\=\int _2^5\:\sqrt{1+x^2-\dfrac{1}{2}+\dfrac{1}{(4x)^2}}dx\\
=\int _2^5\:\sqrt{x^2+\dfrac{1}{2}+\dfrac{1}{(4x)^2}}dx\\
\\=\int _2^5\:\sqrt{\left(x+\frac{1}{4x}\right)^2}\\
=\int _2^5\:|x+\frac{1}{4x}|dx \to (2<x<5) \\
\int _2^5\:(x+\frac{1}{4x})dx$$
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