Question

homework9: problem 2
(1 point)
find the most general antiderivative for the function $\frac{6}{\sqrt3{x}} - 5\sqrt3{x^{2}}$.
antiderivative $=\square$
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Explanation:

Step1: Rewrite the function

First, rewrite the given function using exponent rules. Recall that \(\frac{1}{\sqrt[3]{x}} = x^{-\frac{1}{3}}\) and \(\sqrt[3]{x^2}=x^{\frac{2}{3}}\). So the function \(f(x)=\frac{6}{\sqrt[3]{x}} - 5\sqrt[3]{x^2}\) can be rewritten as \(f(x) = 6x^{-\frac{1}{3}}-5x^{\frac{2}{3}}\).

Step2: Find the antiderivative

To find the antiderivative, we use the power rule for integration, which states that \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C\) (where \(n
eq - 1\) and \(C\) is the constant of integration).

For the first term \(6x^{-\frac{1}{3}}\):
Using the power rule, \(\int6x^{-\frac{1}{3}}dx=6\times\frac{x^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}+C_1=6\times\frac{x^{\frac{2}{3}}}{\frac{2}{3}}+C_1\)
Simplify \(6\times\frac{3}{2}x^{\frac{2}{3}}=9x^{\frac{2}{3}}\)

For the second term \(- 5x^{\frac{2}{3}}\):
Using the power rule, \(\int-5x^{\frac{2}{3}}dx=-5\times\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}+C_2=-5\times\frac{x^{\frac{5}{3}}}{\frac{5}{3}}+C_2\)
Simplify \(-5\times\frac{3}{5}x^{\frac{5}{3}}=-3x^{\frac{5}{3}}\)

Step3: Combine the terms

The most general antiderivative \(F(x)\) is the sum of the antiderivatives of the two terms plus the constant of integration \(C\) (where \(C = C_1 + C_2\)):
\(F(x)=9x^{\frac{2}{3}}-3x^{\frac{5}{3}}+C\)

We can also rewrite the exponents as radicals: \(x^{\frac{2}{3}}=\sqrt[3]{x^2}\) and \(x^{\frac{5}{3}}=\sqrt[3]{x^5}=x\sqrt[3]{x^2}\), so \(F(x) = 9\sqrt[3]{x^2}-3x\sqrt[3]{x^2}+C\) or in the exponent form \(9x^{\frac{2}{3}}-3x^{\frac{5}{3}}+C\)

Answer:

\(9x^{\frac{2}{3}}-3x^{\frac{5}{3}}+C\) (or equivalent forms like \(9\sqrt[3]{x^2}-3x\sqrt[3]{x^2}+C\))