$$\int _0^{\infty }\:\left(\frac{2x}{x^2+5}\right)-\left(\frac{6}{3x+2}\right)$ $ $$\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)$ $
Entonces, $$\ln(\infty ) - \ln(5) - (\ln(∞)−\ln(4)) = \ln(\frac{4}{5})$ $
