Problem 33 - Calculus Early Transcendentals

Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int\left(\frac{3}{x^{4}}+2-\frac{3}{x^{2}}\right) d x$$

Step-by-Step Solution

Answer

Question: Calculate the indefinite integral of the following function: \(f(x) = \frac{3}{x^4} + 2 - \frac{3}{x^2}\). Answer: The indefinite integral of the given function is \(\int\left(\frac{3}{x^{4}}+2-\frac{3}{x^{2}}\right) d x = \frac{-x^{-3}}{1} + 2x -3x^{-1} + C\).

Step 1: Integrate each term of the function separately

Since the integrand is a sum of different functions, we can integrate each term separately: $$\int\left(\frac{3}{x^{4}}+2-\frac{3}{x^{2}}\right) d x = \int\frac{3}{x^{4}} dx + \int2 dx - \int\frac{3}{x^{2}} dx$$

Step 2: Integrate the term 3/x^4

To integrate \(\frac{3}{x^4}\), we can rewrite it as \(x^{-4}\) and proceed with integration: $$\int\frac{3}{x^4}dx = \int3x^{-4}dx$$ Now we can apply the power rule for integration: $$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$ So, for the first term: $$\int3x^{-4}dx = \frac{3x^{-3}}{-3} + C_1$$

Step 3: Integrate the term 2

The integration of a constant term 2 is simply 2 times x plus a constant: $$\int2 dx = 2x + C_2$$

Step 4: Integrate the term -3/x^2

To integrate \(\frac{-3}{x^2}\), we can rewrite it as \(-3x^{-2}\) and proceed with integration: $$\int\frac{-3}{x^2}dx = \int-3x^{-2}dx$$ Now we apply the power rule for integration again: $$\int-3x^{-2}dx = \frac{-3x^{-1}}{-1} + C_3$$

Step 5: Combine the integrated terms

Now, we can combine the antiderivatives from Steps 2, 3, and 4: $$\int\left(\frac{3}{x^{4}}+2-\frac{3}{x^{2}}\right) d x = \frac{3x^{-3}}{-3} + C_1 + 2x + C_2 + \frac{-3x^{-1}}{-1} + C_3$$ This simplifies to: $$\int\left(\frac{3}{x^{4}}+2-\frac{3}{x^{2}}\right) d x = \frac{-x^{-3}}{1} + 2x -3x^{-1} + C$$ Where \(C = C_1 + C_2 + C_3\) is the constant of integration.

Step 6: Check the result by differentiation

To verify our work, we'll differentiate the calculated indefinite integral and check whether it's equal to the original function: $$\frac{d}{dx}\left(\frac{-x^{-3}}{1} + 2x -3x^{-1} + C\right)$$ Using the power rule for differentiation, we obtain: $$\frac{d}{dx}\left(\frac{-x^{-3}}{1} + 2x -3x^{-1} + C\right) = 3x^{-4} + 2 -3x^{-2}$$ This matches the original function, confirming that our solution is correct: $$\int\left(\frac{3}{x^{4}}+2-\frac{3}{x^{2}}\right) d x = \frac{-x^{-3}}{1} + 2x -3x^{-1} + C$$