Question

a. Evaluate \(\int \frac{x^{2}}{x+1} d x\) using the substitution \(u=x+1\) b. Evaluate \(\int \frac{x^{2}}{x+1} d x\) after first performing long division on \(\frac{x^{2}}{x+1}\) c. Reconcile the results in parts (a) and (b).

Short answer

The results for part (a) and part (b) are as follows: Part (a): \(\frac{1}{2}x^2 + x - \frac{3}{2} + \ln|x+1| + C\) Part (b): \(\frac{1}{2}x^2 - x + 2\ln |x+1| + C_4\) These two results are compatible, as they are the same up to a constant, which is expected when comparing two different antiderivatives of a function.

Step-by-step solution

Part (a): Using substitution method

We are given the integral \(\int \frac{x^2}{x+1} dx\) and the substitution \(u=x+1\). Let's first perform the substitution: 1. Substitute \(u = x+1\): We get \(x= u-1\). 2. Differentiate with respect to \(x\): \(dx = du\) as the derivative of \(u\) with respect to \(x\) is 1. Now, substitute the \(x\) and \(dx\) in the integral: \(\int \frac{x^2}{x+1} dx = \int \frac{(u-1)^2}{u} du\) Let's integrate this new integral:

Part (a.1): Expand the integrand

Expand the numerator: \(\int \frac{(u-1)^2}{u} du = \int \frac{u^2-2u+1}{u} du\)

Part (a.2): Separate the integrand

Separate the integrand into individual terms: \(\int \frac{u^2-2u+1}{u} du = \int udu - 2\int 1du + \int \frac{1}{u} du\)

Part (a.3): Integrate each term

Integrate each term: \(\int udu = \frac{1}{2}u^2+C_1\) \(- 2\int 1du = -2u+C_2\) \(\int \frac{1}{u} du = \ln |u| + C_3\)

Part (a.4): Combine the integration results

Combine the results of integration: \(\frac{1}{2}u^2 -2u + \ln |u| + C_4\) Substitute \(x+1\) for \(u\): \(\frac{1}{2}(x+1)^2 -2(x+1) + \ln |x+1| + C = \frac{1}{2}x^2 + x - \frac{3}{2} + \ln|x+1| + C \) This is the result for part (a).

Part (b): Using long division

First, perform long division on \(\frac{x^2}{x+1}\): \(x^2 = (x+1)(x-1) + 2\) Now, integrate the resulting expression: \(\int \frac{x^2}{x+1} dx = \int \left( x-1 + \frac{2}{x+1} \right) dx\)

Part (b.1): Separate the integrand

Separate the integrand into individual terms: \(\int \left( x-1 + \frac{2}{x+1} \right) dx = \int xdx - \int 1dx + 2\int \frac{1}{x+1} dx\)

Part (b.2): Integrate each term

Integrate each term: \(\int xdx = \frac{1}{2}x^2 + C_1\) \(- \int 1dx = -x + C_2\) \(2\int \frac{1}{x+1} dx = 2\ln |x+1| + C_3\)

Part (b.3): Combine the integration results

Combine the results of integration: \(\frac{1}{2}x^2 - x + 2\ln |x+1| + C_4\) This is the result for part (b).

Part (c): Reconcile the results

The results of part (a) and part (b) are as follows: Part (a): \(\frac{1}{2}x^2 + x - \frac{3}{2} + \ln|x+1| + C\) Part (b): \(\frac{1}{2}x^2 - x + 2\ln |x+1| + C_4\) The two results are, in fact, the same when you take into account the constant of integration. The difference in the constants is due to the particular choice of the constant term when we integrate each part. The results in part (a) and part (b) are the same up to a constant, which is expected when comparing two different antiderivatives of a function.

Key concepts

Integration by Substitution

Integration by substitution is a powerful method used to evaluate integrals. It essentially replaces a complex integral with a simpler one, making it easier to solve. Let's take the example of the integral \( \int \frac{x^2}{x+1} dx \). To apply substitution, we set \( u = x + 1 \), transforming the integral into a different form. Here is how it works:

• Determine the substitution: By letting \( u = x + 1 \), a new variable \( u \) is introduced.
• Find the derivative: Differentiate \( u = x + 1 \) to get \( du = dx \). This shows us how \( dx \) relates to \( du \).
• Express \( x \) in terms of \( u \): Since \( u = x + 1 \), it follows that \( x = u - 1 \).
• Substitute back into the integral: Replace all appearances of \( x \) and \( dx \) in the integral with terms involving \( u \), resulting in the integral \( \int \frac{(u-1)^2}{u} du \).

Now, the integration becomes easier as it involves simpler algebraic manipulations. Expand the numerator and separate the integrand to solve each part individually. This systematic approach makes the substitution method an essential tool, especially when dealing with complex integrals.

Long Division in Integration

When facing an integral involving a rational function, such as \( \int \frac{x^2}{x+1} dx \), long division can simplify the process of integration. This technique involves dividing the polynomial in the numerator by the polynomial in the denominator to rewrite the integrand. Here's how long division works for integration:

• Perform division: Divide \( x^2 \) by \( x+1 \) to break it down into more manageable terms.
• Rewrite the integrand: The division will transform the expression into something like \( x - 1 + \frac{2}{x+1} \), separating it into simpler parts.
• Integrate each term: Once the integrand is rewritten, take the integral of each component. For example, \( \int x dx, \int -1 dx, \) and \( 2 \int \frac{1}{x+1} dx \).

Breaking down a complex function using long division helps in identifying the integral of each term separately. This makes solving integrals that share this form more straightforward. Long division helps to simplify the expression significantly before applying fundamental integration techniques.

Reconciling Integration Results

In mathematics, especially integration, different methods may produce different expressions for the same solution. Reconciling these results involves understanding how these expressions relate. For the integral of \( \frac{x^2}{x+1} \), substitution and long division yielded different-looking results. Here's how they reconcile:

• Compare the forms: The expressions might appear different due to the choice of constants during integration.
• Understand constants of integration: Each technique may introduce a different constant, but the difference should be additive, not multiplicative with \( x\).
• Equivalency up to a constant: Despite the differing expressions like \( \frac{1}{2}x^2 + x - \frac{3}{2} + \ln|x+1| + C \) and \( \frac{1}{2}x^2 - x + 2\ln |x+1| + C_4 \), both results are equal except for the constant. This aligns with the property that integrals are unique up to an additive constant.

Thus, reconciling reinforces why constants are an intrinsic part of integration, showing that the two different results are two perspectives on the same underlying antiderivative. For students, understanding this harmony bolsters their grasp of integration principles.