The curve: $\displaystyle y=\frac{x^2}{2} - \frac{lnx}{4}, 2\leq x \leq4$

my work so far:

$\displaystyle x(t)=\frac{x^2}{2} - \frac{ln}{4}$

$\displaystyle x'(t)=x - \frac{1}{4x}$

$\displaystyle y(t)=t$

$\displaystyle y'(t)=1$

$\displaystyle \int^4_2 \sqrt{1^2 + (x - \frac{1}{4x})^2}dx$

$\displaystyle \int^4_2 \sqrt{1+\frac{9}{16}x^2}dx$

anyways after that mess I am really not sure how to integrate this with u-substitution. If I made a mistake please let me know or some direction if I am right so far would be very helpful!! Thanks a lot!