Evaluate an improper logarithmic integral and a revolution volume

Take $u=\ln(3x)$ and $dv=x^{-3}dx$, so $du=dx/x$ and $v=-1/(2x^2)$. Then the integral is $[-\ln(3x)/(2x^2)]_{1/3}^a+\int_{1/3}^a dx/(2x^3)=9/4-\ln(3a)/(2a^2)-1/(4a^2)$.

Both $1/(4a^2)$ and $\ln(3a)/(2a^2)=(\ln3+\ln a)/(2a^2)$ tend to zero. Thus the improper integral is $9/4$.

The integrand is even, so $V=2\pi\int_0^{\sqrt3/2}[3/(1+4x^2)]^2dx=18\pi\int_0^{\sqrt3/2}dx/(1+4x^2)^2$. With $2x=\tan\theta$, $dx=\tfrac12\sec^2\theta\,d\theta$, the bounds are $0$ and $\pi/3$, and $1+4x^2=\sec^2\theta$. Hence $V=9\pi\int_0^{\pi/3}\cos^2\theta\,d\theta$.

$V=\tfrac{9\pi}2[\theta+\tfrac12\sin2\theta]_0^{\pi/3}=\tfrac{9\pi}2(\pi/3+\sqrt3/4)$.