$=\lim_{a\to 2}\int_a^6(x-2)^{-1/2}dx$

Use integration by parts. Let f = ln(x) and dg = dx.

This is the integral of a derivative, so use the Fundamental Theorem of Calculus.

Use partial fraction decomposition.

$\frac{1}{2x(x-3)}=\frac{a}{2x}+\frac{b}{x-3}$

$a(x-3)+b(2x)=1$

$-3a=1\text{ and }ax+2bx=0x$

$a=-\frac{1}{3}\text{ }b=\frac{1}{6}$

$\int_1^2\Big(-\frac{1}{6x}+\frac{1}{6(x-3)}\Big)dx$