<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] Hint. We have that
$$1+\left(x-\frac{1}{4x}\right)^2=\frac{(4x)^2+16x^4-8x^2+1}{(4x)^2}=\frac{16x^4+8x^2+1}{(4x)^2}=\left(\frac{4x^2+1}{4x}\right)^2.$$
</s>