Problem 46 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the indefinite integral. $$\int \frac{t}{t+1} d t$$

Step-by-Step Solution

Answer

The short answer for the indefinite integral of \(\frac{t}{t+1}\) with respect to \(t\) is: \[ \int \frac{t}{t+1} dt = (t + 1) - \ln(|t + 1|) + C \]

Step 1: Choose a substitution

Let us choose the substitution \(u = t+1\). This will simplify the integral and allow us to find the indefinite integral.

Step 2: Differentiate the substitution

We now need to find the differential of our substitution. Differentiating \(u\) with respect to \(t\), we get: \[du = dt\]

Step 3: Integration by substitution

Now, using the substitution \(u = t + 1\) and the fact that \(du = dt\), we can rewrite the integral as follows: \[ \int \frac{t}{t+1} dt = \int \frac{u-1}{u}du \]

Step 4: Simplify the integral

We can simplify this integral as a sum of two simpler integrals: \[ \int \frac{u-1}{u}du = \int \left(\frac{u}{u} - \frac{1}{u}\right)du = \int (1 - u^{-1})du \]

Step 5: Find the indefinite integral

Now, we can find the indefinite integral of the simplified expression: \[ \int (1 - u^{-1})du = u - \int u^{-1}du \] For the second integral, we will use the fact that the integral of \(u^{-1}\) is ln(|u|). So, we get: \[ \int u^{-1}du = \ln(|u|) + C_1 \]

Step 6: Combine the integrals

Combining these two integrals, we get the indefinite integral of the original expression: \[ \int \frac{t}{t+1} dt = u - \ln(|u|) + C \]

Step 7: Substitute back to the original variable

Finally, we replace the \(u\) with our original substitution, which was \(u = t + 1\): \[ \int \frac{t}{t+1} dt = (t + 1) - \ln(|t + 1|) + C \] So, the indefinite integral of \(\frac{t}{t+1}\) with respect to \(t\) is: \[ \int \frac{t}{t+1} dt = (t + 1) - \ln(|t + 1|) + C \]

Key Concepts

Indefinite Integrals

An indefinite integral is fundamentally the opposite operation of differentiation. In calculus, this concept is often referred to as an antiderivative. Given a function, finding its indefinite integral means determining a function whose derivative equals the original function. In notation, the indefinite integral of a function \( f(t) \) is represented by: \[ \int f(t) \, dt \] where \( C \) is the constant of integration. This symbol \( \int \), known as the integral sign, indicates the operation of integration. The constant \( C \) appears because of the indefinite nature of the integral, representing the fact that an infinite number of antiderivatives exist, differing only by a constant.
When you work on indefinite integrals, you're essentially reversing the differentiation process. This means for each derivative that reduces a polynomial by one degree, an integral will increase a polynomial's degree by one.

• For example, if differentiating \( x^n \) results in \( nx^{n-1} \), integrating \( nx^{n-1} \) should return \( x^n \), plus the constant \( C \).
• The integral of \( t \) would be \( \frac{t^2}{2} + C \).

Substitution Method

The substitution method is an integration technique often used to simplify an integral by substituting part of the integral with a new variable. The goal is to transform a complex integral into a simpler one by reorganizing and reformatting the original problem.
The basic steps for using the substitution method are:

• Identify a portion of the integral that can be substituted, typically an expression whose derivative also appears in the integral.
• Re-express the integral in terms of the new variable, \( u \). This usually involves finding \( du \), the differential of the substitution.
• Solve the simplified integral. Often, this new form of the integral is more straightforward to handle.
• Finally, revert to the original variables if necessary.

In the given problem, the substitution of \( u = t + 1 \) shifts the focus from a complex fraction to a more manageable form. Here, \( du = dt \) simplifies the integration, allowing for easier computations.

Integration Techniques

Integration techniques are various methods used to solve integrals, both definite and indefinite, in calculus. Different techniques work best for different types of functions, making it essential to consider which method suits your problem. Here are some common techniques:

• Substitution: A frequently used technique where you change variables to simplify the integral, as covered previously with the substitution method.
• Integration by Parts: Useful for products of functions, it is based on the product rule for differentiation.
• Partial Fractions: Decomposing a complex rational expression into simpler parts to aid in integration.
• Trigonometric Identities: These simplify integrals involving trigonometric functions.

In the exercise, the substitution method was the most fitting choice due to the form of \( \frac{t}{t+1} \). Adjusting the variable helped convert the initial problem into two simpler integrals to handle: \( \int 1 \, du \) and \( \int u^{-1} \, du \). After integrating, solutions like \( u - \int u^{-1} \, du \) make the problem more approachable, demonstrating the effectiveness of choosing the right technique.
It's about clever manipulation and recognizing patterns to apply the best technique.