Distinguish a logarithm from its absolute-value graph

$\int\ln(2x-1)dx=x\ln(2x-1)-\int2x/(2x-1)dx=x\ln(2x-1)-x-\tfrac12\ln(2x-1)+C=(x-\tfrac12)\ln(2x-1)-x+C$, for $x>1/2$.

The absolute-value graph has domain $x>1/2$ and vertical asymptote $x=1/2$. It decreases to the corner $(1,0)$, then increases; there is no y-intercept. For $1/2<x<1$ it is $-\ln(2x-1)$; for $x\ge1$ it is $\ln(2x-1)$. The rendered source has modulus bars in (ii) and (iii), which extraction omits; (iv) uses the ordinary positive logarithm on its stated interval.

Writing $F(x)=(x-\tfrac12)\ln(2x-1)-x$, the area is $-[F]_{3/4}^1+[F]_1^{3/2}=\tfrac34\ln2-\tfrac14$ square units.

At height $y$, the curve has $x=(e^y+1)/2$. The source setup is $V=9\pi\ln5-(9\pi/4)\ln2-\pi\int_{\ln2}^{\ln5}[(e^y+1)/2]^2dy$. Equivalently shells give $2\pi\int_{3/2}^3x\ln(2x-1)dx$. Evaluation gives $\pi[\tfrac{35}4\ln5-2\ln2-\tfrac{33}8]\approx26.9$ cubic units (3 s.f.).