Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-54, perform the indicated integrations. \(\int \frac{x^{3}+7 x}{x-1} d x\)

Step-by-Step Solution

Answer

The integral is \(\frac{x^3}{3} + \frac{x^2}{2} + 8x + 8 \ln |x-1| + C\).

Step 1: Polynomial Long Division

Before integrating, we need to simplify the given rational function \(\frac{x^3 + 7x}{x-1}\) by using polynomial long division. We divide \(x^3 + 7x\) by \(x-1\):1. Divide \(x^3\) by \(x\), which gives \(x^2\).2. Multiply \(x^2\) by \(x-1\) to get \(x^3 - x^2\), subtract this from the original divisor to get \(x^2 + 7x\).3. Divide \(x^2\) by \(x\) to get \(x\).4. Multiply \(x\) by \(x-1\) to get \(x^2 - x\), subtract to get \(8x\).5. Divide \(8x\) by \(x\) to get \(8\).6. Multiply \(8\) by \(x-1\) to get \(8x - 8\), subtract to get the remainder \(8\).Thus, \(\frac{x^3 + 7x}{x-1} = x^2 + x + 8 + \frac{8}{x-1}\).

Step 2: Break Down the Integral

The integral now becomes: \[ \int (x^2 + x + 8 + \frac{8}{x-1}) \, dx \]This integral can be separated into: \[ \int x^2 \, dx + \int x \, dx + \int 8 \, dx + \int \frac{8}{x-1} \, dx \].

Step 3: Solve Each Integral

Let's integrate each part separately:1. \(\int x^2 \, dx\) results in \(\frac{x^3}{3} + C_1\).2. \(\int x \, dx\) results in \(\frac{x^2}{2} + C_2\).3. \(\int 8 \, dx\) results in \(8x + C_3\).4. \(\int \frac{8}{x-1} \, dx\) results in \(8 \ln |x-1| + C_4\).These constants \(C_1, C_2, C_3, C_4\) combine into a single constant \(C\).

Step 4: Combine the Results

Now, combine the results of the integrals:\[ \frac{x^3}{3} + \frac{x^2}{2} + 8x + 8 \ln |x-1| + C \]This is the antiderivative of the given function.

Key Concepts

Polynomial Long Division

Polynomial long division is a technique used to divide a polynomial by a simpler polynomial, just like how arithmetic long division works with numbers. It comes into play when dealing with the division of two polynomials that don't easily factor or cancel out. In our exercise, we started with the polynomial \[ \frac{x^3 + 7x}{x-1} \]- Identify the dividend (\[x^3 + 7x \]) and the divisor (\[x-1\]).- To begin, find out how many times the leading term in the divisor (\[x\]) can multiply into the leading term of the dividend (\[x^3\]). Here, it goes \[x^2\] times.- Multiply and subtract to find the difference, then repeat the process with the new dividend. - Keep reducing: Go from \[x^3 - x^2\] to \[x^2 + 7x\], and so on.- Continue until dividing the terms is no longer possible, and the remainder shrinks smaller than the degree of the divisor.The result of performing polynomial long division on our rational function is \[ x^2 + x + 8 + \frac{8}{x-1} \].

Rational Functions

Rational functions are expressions formed by the ratio of two polynomials. They can often be complex to integrate directly, which is why simplification techniques, like polynomial long division, are used.- **General Form**: Given as \[ \frac{p(x)}{q(x)} \], where both \(p(x)\) and \(q(x)\) are polynomials.- In our example, the rational function is \[ \frac{x^3 + 7x}{x-1} \].- By simplifying it as \[ x^2 + x + 8 + \frac{8}{x-1} \], we get a mixture of polynomial and simpler rational terms.Why simplify? Well, dividing and simplifying can make integration more straightforward and feasible, especially when the numerator has a higher degree than the denominator.

Integral Calculus

Integral calculus focuses on finding antiderivatives and calculating areas under curves. When you see a function and a \( \int \), it means you're dealing with integration.To integrate our broken down rational function, consider each component separately:- **Separating Terms:** \[ \int (x^2 + x + 8 + \frac{8}{x-1}) \, dx \]- This simplifies to the integration of each term: - \( \int x^2 \, dx \) - \( \int x \, dx \) - \( \int 8 \, dx \) - \( \int \frac{8}{x-1} \, dx \)The goal here is to address each part individually to make the overall integration easier.

Antiderivatives

Antiderivatives are functions that represent the 'reverse' process of differentiation. When we integrate a function, what we're acquiring is essentially an antiderivative.- **Using Basic Rules:** - Integrate polynomial terms to find their antiderivatives using power rules: - \( \int x^2 \, dx = \frac{x^3}{3} + C \) - \( \int x \, dx = \frac{x^2}{2} + C \) - \( \int 8 \, dx = 8x + C \)- **Rational Integration:** - For terms like \( \int \frac{8}{x-1} \, dx \), use the natural logarithm rule: - Result is \( 8 \ln |x-1| + C \)In summary, the antiderivative gives us a general solution represented with a constant \( C \), capturing all possible shifts of our solution along the y-axis. The proper joining of these individual integrals results in the complete antiderivative of our original function.