Let $ f(x) = \int \frac{dx}{x^{2/3} + 2x^{1/2}} $ be such that $ f(0) = -26 + 24\log_e 2 $. If $ f(1) = a + b\log_e 3 $, where $ a, b \in \mathbb{Z} $, then $ a + b $ is equal to

$ f(x) = \int \frac{dx}{x^{2/3} + 2x^{1/2}} $

Put $ x = t^6 \Rightarrow dx = 6t^5 dt $

$ = \int \frac{6t^5 dt}{t^4 + 2t^3} = 6\int \frac{t^2 - 4}{t + 2} dt $

$ = 6\left[ \frac{t^2}{2} - 2t + 4\ln(t + 2) \right] + C $

$ = 3x^{1/3} - 12x^{1/6} + 24\ln(x^{1/6} + 2) + C $

$ f(0) = 24\ln 2 + C = -26 + 24\ln 2 \Rightarrow C = -26 $

Now

$ f(1) = -35 + 24\ln 3 = a + b\ln 3 $

$ \Rightarrow a = -35,; b = 24 $

$ \Rightarrow a + b = -11 $