Let f(x) = int frac{dx}{x^{2/3} + 2x^{1/2}} b | vedclass

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(C) Given $f(x) = \int \frac{dx}{x^{2/3} + 2x^{1/2}}$.Let $x = t^6$, then $dx = 6t^5 dt$.Substituting these into the integral:$f(x) = \int \frac{6t^5

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$-11$

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(C) Given $f(x) = \int \frac{dx}{x^{2/3} + 2x^{1/2}}$.Let $x = t^6$, then $dx = 6t^5 dt$.Substituting these into the integral:$f(x) = \int \frac{6t^5 dt}{t^4 + 2t^3} = \int \frac{6t^2 dt}{t + 2} = 6 \int \frac{t^2 - 4 + 4}{t + 2} dt$.$f(x) = 6 \int (t - 2 + \frac{4}{t + 2}) dt = 6 [\frac{t^2}{2} - 2t + 4 \log_{e}(t + 2)] + C$.Substituting $t = x^{1/6}$:$f(x) = 3x^{1/3} - 12x^{1/6} + 24 \log_{e}(x^{1/6} + 2) + C$.Given $f(0) = -26 + 24 \log_{e}(2)$.At $x = 0$, $f(0) = 0 - 0 + 24 \log_{e}(2) + C = 24 \log_{e}(2) + C$.Comparing, $C = -26$.Now, $f(1) = 3(1)^{1/3} - 12(1)^{1/6} + 24 \log_{e}(1^{1/6} + 2) - 26$.$f(1) = 3 - 12 + 24 \log_{e}(3) - 26 = -35 + 24 \log_{e}(3)$.Given $f(1) = a + b \log_{e}(3)$, we have $a = -35$ and $b = 24$.Therefore, $a + b = -35 + 24 = -11$.

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:

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