Why won't maxima evaluate this indefinite integral

Question

The integral in question is $$\int {\sqrt{\left({{x}\over{2}}-{{1}\over{2\,x}}\right)^2+1}}\ dx.$$ I need the intermediate step in determining the arc length of a function from $[1, 2]$. Maxima will give me the answer $\left(\frac{\log 2}{2}+{{3}\over{4}}\right)$, but not the indefinite integral. Why is this happening and is there a way for me to get around this?

Answer 1

We have $$\int {\sqrt{\left({{x}\over{2}}-{{1}\over{2\,x}}\right)^2+1}}\ dx$$ $$=\frac{1}{2}\int\sqrt{(x-\frac{1}{x})^2+4} \space dx$$ $$=\frac{1}{2}\int\sqrt{x^2-2+\frac{1}{x^2}+4} \space dx$$ $$=\frac{1}{2}\int\sqrt{\big(\frac{x^2+1}{x}\big)^2} \space dx$$ $$=\frac{1}{2} \int\frac{x^2+1}{x}dx$$ $$=\frac{1}{2}\big[\int x dx+\int \frac{1}{x} dx \big]+C$$ $$=\frac{x^2+2\ln(|x|)}{4}+C.$$

Thus $$\int_{1}^{2} {\sqrt{\left({{x}\over{2}}-{{1}\over{2\,x}}\right)^2+1}}\ dx$$ $$=\frac{4+2\ln(2)}{4}-\frac{1}{4}=\frac{\ln{2}}{2}+\frac{3}{4}.$$