Problem 55 - Calculus Early Transcendentals

Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int \frac{t+1}{t} d t$$

Step-by-Step Solution

Answer

Answer: The indefinite integral of the function $\frac{t+1}{t}$ is $t + \ln |t| + C$, where C is the constant of integration.

Step 1: Simplify the Function

Divide t into both terms in the numerator to simplify the given function: $$\frac{t+1}{t} = \frac{t}{t} + \frac{1}{t} = 1 + \frac{1}{t}$$ Now, we will integrate each term separately.

Step 2: Apply the Indefinite Integral

Integrate each term with respect to t: $$\int \left(1 + \frac{1}{t}\right) dt = \int 1 dt + \int \frac{1}{t} dt$$

Step 3: Integrate the Terms

Integrate each term separately: $$\int 1 dt = t$$ $$\int \frac{1}{t} dt = \ln |t|$$ Now combine both of the integrals: $$t + \ln |t| + C$$ where C is the constant of integration. So, the indefinite integral of the given function is: $$\int \frac{t+1}{t} d t = t + \ln |t| + C$$

Step 4: Check Your Work by Differentiating

Now, differentiate the calculated integral to ensure that it matches the original function: $$\frac{d}{dt}\left(t + \ln |t| + C\right) = \frac{d}{dt}(t) + \frac{d}{dt}(\ln |t|) + \frac{d}{dt}(C)$$ Differentiate each term: $$\frac{d}{dt}(t) = 1$$ $$\frac{d}{dt}(\ln |t|) = \frac{1}{t}$$ $$\frac{d}{dt}(C) = 0$$ Combine these derived terms to get the original function: $$1 + \frac{1}{t} = \frac{t+1}{t}$$ As we can see, the derivative of the calculated integral matches the original function, confirming that the indefinite integral was found correctly: $$\int \frac{t+1}{t} d t = t + \ln |t| + C$$