Problem 51 - College Algebra and Calculus: An Applied Approach

Question

Use any basic integration formula or formulas to find the indefinite integral. State which integration formula(s) you used to find the integral. $$ \int \frac{x^{2}+2 x+5}{x-1} d x $$

Step-by-Step Solution

Answer

The integral of \(\frac{x^{2}+2 x+5}{x-1}\) is \(\frac{1}{2}x^{2}+ 3x + 8\ln|x-1| + C\).

Step 1: Rewrite the numerator

Rewrite the numerator in the form \((x-1)Q(x) + R(x)\) where \(Q(x)\) is the quotient and \(R(x)\) is the remainder when divided by \(x-1\). In this case, we can rewrite the numerator as \(x^{2}+2 x+5 = (x - 1)\cdot(x + 3) + 8\). Thus, the integral becomes \(\int \frac{(x - 1)\cdot(x + 3) + 8}{x-1} d x\).

Step 2: Split the integral

Split the integral into two: \(\int (x + 3) d x + \int \frac{8}{x-1} d x\). The first integral is a simple polynomial, and the second is a logarithmic function.

Step 3: Integrate the terms

Integrate the two terms: The integral of \(x + 3\) is \(\frac{1}{2}x^{2}+ 3x + C\) (applying power rule of integration) and the integral of \(\frac{8}{x-1}\) is \(8\ln|x-1|\) (using the formula \(\int \frac{1}{x} d x = \ln|x| + C\). Therefore, the integral of the expression is \(\frac{1}{2}x^{2}+ 3x + 8\ln|x-1| + C\).

Key Concepts

Basic Integration Formulas

When it comes to integrating functions, having an arsenal of basic integration formulas at your disposal can be tremendously helpful. Key integration formulas serve as the foundation for solving more complex integrals. Here are a few essentials you should know:

• Power Rule: This formula is used to integrate functions of the form \(x^n\), where \(n eq -1\). The rule states: \[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]


• Summation Rule: This allows for the integration of the sum of functions separately: \[\int [f(x) + g(x)] \, dx = \int f(x) \, dx + \int g(x) \, dx\]


• Logarithmic Rule: This rule is especially useful when integrating expressions of the form \(\frac{1}{x}\). It is stated as:\[\int \frac{1}{x} \, dx = \ln|x| + C\]

These basic rules provide the scaffolding for approaching indefinite integrals, making them manageable, whether it's a simple polynomial or a logarithmic expression.

Polynomial Integration

Polynomial integration is a relatively straightforward process because it leverages the Power Rule of integration. When given a polynomial function, each term can often be integrated individually. For instance, consider the polynomial expression in the exercise:\[x + 3\]Each term in this polynomial is integrated separately. Let’s break it down further:

• For the term \(x\), using the Power Rule:\[\int x \, dx = \frac{x^2}{2} + C\]


• For the constant term \(3\):\[\int 3 \, dx = 3x + C\]

When combined, the integral of \(x + 3\) is:\[\frac{1}{2}x^2 + 3x + C\]Understanding polynomial integration is critical because it shows the simplicity of the Power Rule when dealing with straight-forward polynomial expressions.

Logarithmic Integration

Logarithmic integration is necessary when the integrand includes a rational expression where the degree of the numerator is less than that of the denominator, typically in the form of \(\frac{1}{x}\) or more generally \(\frac{u'}{u}\). In our exercise, one of the integrals derived is:\[\int \frac{8}{x-1} \, dx\]This falls exactly into the realm of logarithmic integration. Here's how you would handle this:

• Firstly, recognize that this expression can be integrated using the logarithmic rule:\[\int \frac{1}{u} \, du = \ln|u| + C\]


• In our integral, let \(u = x-1\). The derivative \(u'\) is thus 1, simplifying our expression to:\[8\int \frac{1}{x-1} \, dx = 8\ln|x-1| + C\]

Working with logarithmic integration requires recognition of the form \(\frac{1}{u}\) and simplifies seemingly complex rational integrals into manageable steps, leading to solutions involving natural logarithms.