Problem 29 - Calculus Early Transcendentals

Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int\left(3 x^{1 / 3}+4 x^{-1 / 3}+6\right) d x$$

Step-by-Step Solution

Answer

Question: Find the indefinite integral of the function \(3x^{1/3}+4x^{-1/3}+6\). Answer: The indefinite integral of the given function is \(\frac{9}{4}x^{4/3} + 6x^{2/3} + 6x + C\).

Step 1: Integrate each term individually

We have the integral: $$\int\left(3 x^{1 / 3}+4 x^{-1 / 3}+6\right) d x$$ We will integrate each term individually: 1. \(\int 3x^{1/3} dx\) 2. \(\int 4x^{-1/3} dx\) 3. \(\int 6 dx\)

Step 2: Apply the power rule

Apply the power rule of integration to each term: 1. $$\int 3x^{1/3} dx = 3\int x^{1/3} dx = 3 \cdot \frac{x^{1/3 + 1}}{1/3 + 1} + C_1 = \frac{9}{4}x^{4/3} + C_1$$ 2. $$\int 4x^{-1/3} dx = 4\int x^{-1/3} dx = 4 \cdot \frac{x^{-1/3 + 1}}{-1/3 + 1} + C_2 = 6x^{2/3} + C_2$$ 3. $$\int 6 dx = 6x + C_3$$

Step 3: Combine all the results

Combine the results from step 2: $$\int\left(3 x^{1 / 3}+4 x^{-1 / 3}+6\right) d x = \frac{9}{4}x^{4/3} + 6x^{2/3} + 6x + C$$ where \(C = C_1 + C_2 + C_3\).

Step 4: Check the result by differentiation

Check the result by finding the derivative of our solution: $$\frac{d}{dx}\left(\frac{9}{4}x^{4/3} + 6x^{2/3} + 6x + C\right)$$ Using the power rule of differentiation: $$=3 x^{1/3} + \frac{12}{3}x^{-1/3} + 6$$ This matches the original function, confirming our integration is correct.