Arclength, finding definite integral

Question

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.

Answer 1

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

Answer 2

$$L = \int _2^5\:\sqrt{1+\left(f'\left(x\right)^2\right)} \\ \\=\int _2^5\:\sqrt{1+\left(x-\frac{1}{4x}\right)^2}\\ \\=\int _2^5\:\sqrt{1+x^2-\dfrac{1}{2}+\dfrac{1}{(4x)^2}}dx\\ =\int _2^5\:\sqrt{x^2+\dfrac{1}{2}+\dfrac{1}{(4x)^2}}dx\\ \\=\int _2^5\:\sqrt{\left(x+\frac{1}{4x}\right)^2}\\ =\int _2^5\:|x+\frac{1}{4x}|dx \to (2<x<5) \\ \int _2^5\:(x+\frac{1}{4x})dx$$