Indefinite Integrals

Concept:Integration by substitution is used to simplify the integrand.We choose $x = t^6$ to convert fractional exponents into integer powers.Then integrate term by term and apply the given initial condition to find the constant.Explanation:Rewrite the integral: $f(x) = \int \frac{dx}{x^{2/3} + 2x^{1/2}}$.Substitute $x = t^6$ so that $dx = 6t^5 dt$.Then $x^{2/3} = (t^6)^{2/3} = t^4$ and $x^{1/2} = (t^6)^{1/2} = t^3$.The integral becomes: $\int \frac{6t^5}{t^4 + 2t^3} dt = \int \frac{6t^2}{t+2} dt$.Divide: $\frac{t^2}{t+2} = \frac{(t^2-4)+4}{t+2} = (t-2) + \frac{4}{t+2}$.Integrate: $f(x) = 6 \int \left( t - 2 + \frac{4}{t+2} \right) dt = 6 \left( \frac{t^2}{2} - 2t + 4\ln|t+2| \right) + C$.Simplify: $f(x) = 3t^2 - 12t + 24\ln|t+2| + C$.Back substitute $t = x^{1/6}$: $f(x) = 3x^{1/3} - 12x^{1/6} + 24\ln|x^{1/6} + 2| + C$.Use $f(0) = -26 + 24\ln 2$: $f(0) = 3(0) - 12(0) + 24\ln(0+2) + C = 24\ln 2 + C$.Thus $24\ln 2 + C = -26 + 24\ln 2$, so $C = -26$.Now $f(1) = 3(1)^{1/3} - 12(1)^{1/6} + 24\ln(1^{1/6} + 2) - 26 = 3 - 12 + 24\ln 3 - 26 = -35 + 24\ln 3$.Comparing with $f(1) = a + b\ln 3$, we get $a = -35$, $b = 24$.Thus $a + b = -35 + 24 = -11$.Answer:$a + b = -11$ (Option B).