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

Given the integral:

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

and the condition $$f(0) = -26 + 24 \log_e 2$$, we need to find $$f(1) = a + b \log_e 3$$ and then $$a + b$$.

To solve the integral, use the substitution $$t = x^{\frac{1}{6}}$$. Then:

$$x = t^6, \quad dx = 6t^5 dt$$

Rewrite the exponents:

$$x^{\frac{2}{3}} = (t^6)^{\frac{2}{3}} = t^4, \quad x^{\frac{1}{2}} = (t^6)^{\frac{1}{2}} = t^3$$

Substitute into the integral:

$$f(x) = \int \frac{6t^5}{t^4 + 2t^3} dt = \int \frac{6t^5}{t^3(t + 2)} dt = \int \frac{6t^2}{t + 2} dt$$

Simplify $$\frac{t^2}{t + 2}$$ by polynomial division:

$$\frac{t^2}{t + 2} = t - 2 + \frac{4}{t + 2}$$

Thus:

$$\int \frac{6t^2}{t + 2} dt = 6 \int \left( t - 2 + \frac{4}{t + 2} \right) dt = 6 \left( \frac{t^2}{2} - 2t + 4 \log_e |t + 2| \right) + C$$

$$= 3t^2 - 12t + 24 \log_e |t + 2| + C$$

Substitute back $$t = x^{\frac{1}{6}}$$:

$$t^2 = (x^{\frac{1}{6}})^2 = x^{\frac{1}{3}}, \quad t = x^{\frac{1}{6}}$$

So:

$$f(x) = 3x^{\frac{1}{3}} - 12x^{\frac{1}{6}} + 24 \log_e |x^{\frac{1}{6}} + 2| + C$$

For $$x > 0$$, $$x^{\frac{1}{6}} + 2 > 0$$, so the absolute value can be removed:

$$f(x) = 3x^{\frac{1}{3}} - 12x^{\frac{1}{6}} + 24 \log_e (x^{\frac{1}{6}} + 2) + C$$

Given $$f(0) = -26 + 24 \log_e 2$$, interpret $$f(0)$$ as the limit as $$x \to 0^+$$:

$$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left[ 3x^{\frac{1}{3}} - 12x^{\frac{1}{6}} + 24 \log_e (x^{\frac{1}{6}} + 2) + C \right] = 0 - 0 + 24 \log_e (2) + C$$

Set equal to the given value:

$$24 \log_e 2 + C = -26 + 24 \log_e 2 \implies C = -26$$

Thus:

$$f(x) = 3x^{\frac{1}{3}} - 12x^{\frac{1}{6}} + 24 \log_e (x^{\frac{1}{6}} + 2) - 26$$

Now evaluate $$f(1)$$:

$$f(1) = 3(1)^{\frac{1}{3}} - 12(1)^{\frac{1}{6}} + 24 \log_e (1^{\frac{1}{6}} + 2) - 26 = 3(1) - 12(1) + 24 \log_e (1 + 2) - 26$$

$$= 3 - 12 + 24 \log_e 3 - 26 = -35 + 24 \log_e 3$$

So $$a = -35$$, $$b = 24$$, and:

$$a + b = -35 + 24 = -11$$

The correct option is D. -11.