$\displaystyle y=\frac{x^{3\slash 2}}{3}-x^{1\slash 2}\Longrightarrow y'=\frac{x^{1\slash 2}}{2}-\frac{1}{2x^{1\slash 2}}$
$\displaystyle 1+(y')^2=1+\frac{1}{4}x-\frac{1}{2}+\frac{1}{4x}=\frac{1}{4}\left(x+2+\fra c{1}{x}\right)=\frac{1}{4}\left(x+\frac{1}{x}\righ t)^2$
So the function in the integral is pretty simple: $\displaystyle {1}{2}\left(x+\frac{1}{x}\right)$ ...unfortunately this improper integral diverges at zero since $\displaystyle \ln x\xrightarrow[x\to 0]{}-\infty$ ...
Tonio
No, tonio, that last term is
$\displaystyle \frac{1}{4}\left(x^{1/2}+ \frac{1}{x^{1/2}}\right)^2$.
The integral becomes $\displaystyle \frac{1}{2}\int_0^1 \left(x^{1/2}+ x^{-1/2}\right)dx$ which, while an improper integral, does have a finite value.
So the function in the integral is pretty simple: $\displaystyle {1}{2}\left(x+\frac{1}{x}\right)$ ...unfortunately this improper integral diverges at zero since $\displaystyle \ln x\xrightarrow[x\to 0]{}-\infty$ ...
Tonio