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Chapter 4: Problem 36

Determine the following indefinite integrals. Check your work by differentiation. $$\int \frac{12 t^{8}-t}{t^{3}} d t$$

Short Answer

Expert verified
Answer: The anti-derivative of the given integrand function is: $$\int \frac{12 t^{8}-t}{t^{3}} d t = 2t^6 - \frac{1}{t} + C$$

Step by step solution

01

Break the integrand into terms

Separate the function in the given integral into individual terms. This allows us to integrate each term individually. $$\int \frac{12 t^{8}-t}{t^{3}} d t = \int 12 t^{5} dt - \int \frac{1}{t^2} dt$$
02

Integrate each term

Calculate the indefinite integral of each term: 1. For \(12 t^5\), we apply the power rule of integration: \(\int x^n dx = \frac{x^{n+1}}{n+1}+C\). $$\int 12 t^{5} dt = \frac{12 t^{5+1}}{5+1} + C = 2 t^6 + C_1$$ 2. For \(\frac{1}{t^2}\), we rewrite it as \(t^{-2}\) and apply the power rule of integration: $$\int t^{-2} dt = \frac{t^{-2+1}}{-2+1} + C = -t^{-1} + C_2 = -\frac{1}{t} + C_2$$
03

Combine the integrated terms

Combine the results of the previous step to obtain the indefinite integral of the given function: $$\int \frac{12 t^{8}-t}{t^{3}} d t = 2t^6 - \frac{1}{t} + C$$ where \(C = C_1 + C_2\) is the constant of integration.
04

Verify the result using differentiation

Differentiate the anti-derivative found in step 3 with respect to \(t\) to check if it results in the original integrand function: $$\frac{d\big(2t^6 - \frac{1}{t} + C\big)}{dt} = 12t^5 + \frac{1}{t^2}$$ Since the outcome matches the original integrand in the integral, the result is verified.

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