$$\int _0^{\infty }\:\left(\frac{2x}{x^2+5}\right)-\left(\frac{6}{3x+2}\right)$$ $$\int _0^{\infty }\:\left(\frac{2x}{x^2+5}\right)-2\int _0^{\infty }\:\left(\frac{3}{3x+2}\right)$$ $$\lim _{b\to \infty }\left[\ln\left(x^2+5\right)\right]^{b}_{0} = \ln(\infty ) - \ln(5) $$ $$2\cdot \left(\lim _{b\to \infty }\left[\ln\left(3b+2\right)\right]^{b}_{0}-\ln\left(2\right)\right) = \ln(∞)−\ln(4)$$
So, $$\ln(\infty ) - \ln(5) - (\ln(∞)−\ln(4)) = \ln(\frac{4}{5})$$
Hint: to solve this correctly, notice that $$\ln(x^2+5) - 2\ln|3x+2| = \ln \frac{x^2+5}{(3x+2)^2}$$