Problem 82 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(71-84\) $$ \int \frac{t^{3}-2 t^{2}+3 t-4}{t^{2}+1} d t $$

Step-by-Step Solution

Answer

\[ \frac{t^2}{2} - 2t + \ln|t^2 + 1| - 4 \tan^{-1}t + C \]

Step 1: Polynomial Long Division

We start by dividing the polynomial in the numerator by the polynomial in the denominator. Here we divide \(t^3 - 2t^2 + 3t - 4\) by \(t^2 + 1\) to simplify the expression.1. Divide the leading term \(t^3\) by \(t^2\) to get \(t\).2. Multiply \(t\) by \(t^2 + 1\) to get \(t^3 + t\).3. Subtract \(t^3 + t\) from \(t^3 - 2t^2 + 3t - 4\), resulting in \(-2t^2 + 2t - 4\).4. Repeat the process: divide \(-2t^2\) by \(t^2\) to get \(-2\), and multiply to subtract and get the remainder "\(2t - 4\)".The result of the division is \(t - 2\) with a remainder of \(2t - 4\).

Step 2: Rewriting the Integral

Rewrite the integral into two parts using the result from polynomial long division:\[\int (t - 2 + \frac{2t - 4}{t^2 + 1}) \, dt.\]

Step 3: Evaluating the Integral

Now integrate each term separately:1. Integrate \(t\): \[\int t \, dt = \frac{t^2}{2}.\]2. Integrate \(-2\): \[\int -2 \, dt = -2t.\]3. Integrate \(\frac{2t - 4}{t^2 + 1}\) using substitution. Let \(u = t^2 + 1\), then \(du = 2t \, dt\), leading to: \[\int \frac{2t}{t^2 + 1} \, dt = \int \frac{1}{u} \, du = \ln|u| = \ln|t^2 + 1|.\]4. Integrate the constant term "\(-4\)" with respect to t, which is essentially the same as integrating \(-4 \cdot \frac{1}{t^2+1}\): \[\int -4 \cdot \frac{1}{t^2 + 1} \, dt = -4 \tan^{-1}t.\]

Step 4: Combining the Results

Combine the results of each integration:The integral becomes:\[ \frac{t^2}{2} - 2t + \ln|t^2 + 1| - 4 \tan^{-1}t + C, \]where \(C\) is the constant of integration.

Key Concepts

Polynomial Division

Polynomial division is a technique used to divide a polynomial by another polynomial, similar to long division with numbers. In integral calculus, it helps simplify expressions for easier integration.
To perform polynomial division:

• Divide the leading term of the numerator by the leading term of the denominator.
• Multiply the entire divisor by the result.
• Subtract this from the original dividend to find the remainder.
• Repeat with the next terms, if necessary.

In our exercise, we divided the polynomial \(t^3 - 2t^2 + 3t - 4\) by \(t^2 + 1\), resulting in a quotient of \(t - 2\) and a remainder \(2t - 4\). This simplifies the integration process by breaking it down into more manageable parts.

Integration Techniques

Integration techniques are methods used to solve integrals more easily. By applying specific techniques, we can tackle complex integrals by transforming them into simpler forms.
Here are some common techniques:

• Direct integration of simple functions, like \(t\) or constants.
• Substitution for transforming variables.
• Integration by parts, useful when a product of functions is involved.
• Partial fraction decomposition, which breaks down complex rational functions.

In our example, after polynomial division, the problem is split into smaller integrals. This division allows us to apply straightforward integration for terms like \(t\) and \(-2\), and prepare for substitution for the remaining fraction.

Substitution Method

The substitution method is a powerful integration technique that simplifies integrals by changing the variable. It's particularly useful when dealing with composite functions.For substitution:

• Identify a part of the integrand to set as \(u\). Ideally, the derivative of \(u\) should be present in the integral.
• Replace the identified part with \(u\) and transform the differential. For instance, if \(u = t^2 + 1\), then \(du = 2t \, dt\).
• Integrate with respect to \(u\).
• Finally, substitute back the original variable.

In our problem, the substitution \(u = t^2 + 1\) simplified the integral of \(\frac{2t}{t^2 + 1}\), turning it into \(\int \frac{1}{u} \, du\), which is easier to solve and results in \(\ln|t^2 + 1|\).

Constant of Integration

The constant of integration is a term added when performing indefinite integration. Its role is to account for all potential antiderivatives.When integrating, we find one solution among infinitely many. Each solution differs by a constant:

• Indefinite integrals don't have boundary conditions to fix one specific answer, so we include \(C\).
• This constant \(C\) represents all possible vertical shifts of the antiderivative graph.
• In problems involving specific values, like initial conditions, \(C\) can be determined.

In our solved integral, the result includes \(C\), expressed as \[ \frac{t^2}{2} - 2t + \ln|t^2 + 1| - 4 \tan^{-1}t + C \]. This reminds us of the infinite possibilities beyond the computed expression.