Question

Evaluate \(\int \frac{x-3}{x+2} d x\)

Short answer

The integral evaluates to \( x - 5 \ln|x+2| + C \).

Step-by-step solution

Simplify the Integrand

We begin with the integral \( \int \frac{x-3}{x+2} \, dx \). The first step is to perform polynomial long division to simplify the integrand. Dividing \( x-3 \) by \( x+2 \), we get the quotient \( 1 \) and a remainder of \( -5 \). Thus, \( \frac{x-3}{x+2} = 1 - \frac{5}{x+2} \).

Rewrite the Integral

Given that \( \frac{x-3}{x+2} = 1 - \frac{5}{x+2} \), we rewrite the integral as \( \int \left( 1 - \frac{5}{x+2} \right) \, dx \). This simplifies to two separate integrals: \( \int 1 \, dx - \int \frac{5}{x+2} \, dx \).

Integrate Each Term Separately

Now, integrate each part: the integral of \( 1 \, dx \) is \( x \), and the integral of \( -\frac{5}{x+2} \, dx \) is: \( -5 \int \frac{1}{x+2} \, dx \). The integral of \( \frac{1}{x+2} \, dx \) is \( \ln|x+2| \), so \( -5 \ln|x+2| \).

Write the Final Result

Combine the results from the previous step to get the final output of the integration process: \( x - 5 \ln|x+2| + C \), where \( C \) is the constant of integration. Therefore, the solution to the integral is \( x - 5 \ln|x+2| + C \).

Key concepts

Polynomial Long Division

Polynomial long division is a technique used to divide one polynomial by another, similar to dividing numbers. This process is essential when dealing with complex integral expressions. It helps simplify the given function to a form that's easier to integrate.
To perform polynomial long division, consider the function as a division problem, where the numerator is divided by the denominator:

• Identify the dividend (numerator) and the divisor (denominator). In our exercise, these are \( x-3 \) and \( x+2 \).
• Start by dividing the leading term of the dividend by the leading term of the divisor.
• Multiply the whole divisor by the result and subtract it from the dividend.
• Repeat the process with the new dividend until the remainder's degree is less than the divisor's degree.

In our example, dividing \( x-3 \) by \( x+2 \) yields a quotient of \( 1 \) and a remainder of \( -5 \). Thus, the expression simplifies to \( 1 - \frac{5}{x+2} \), making it more manageable for integration.

Definite and Indefinite Integrals

Integrals are fundamental to calculus, representing areas under curves or the cumulative sum of a function. They come in two main types: definite and indefinite.
Indefinite integrals are denoted by an integral symbol without limits, like \( \int f(x) \, dx \). They represent a family of functions resulting in antiderivatives, plus an arbitrary constant \( C \). In our exercise, the integral \( \int \frac{x-3}{x+2} \, dx \) is indefinite, leading to a general function form as part of its solution.
On the other hand, definite integrals calculate the exact area under the curve between two bounds \( a \) and \( b \). Their results are specific numbers rather than functions, given by \( \int_{a}^{b} f(x) \, dx \).
For indefinite integrals, such as the one in our task, ensuring that each part of the function is integrated correctly is pivotal. Thereby, it involves breaking down the expression into simpler parts, which can then be handled separately and recombined for the final solution.

Natural Logarithm Integration

Natural logarithm integration involves functions of the form \( \ln(x) \), which often appear in integrals, especially when the integrand contains a fraction. When integrating logarithmic functions, the integration of \( \frac{1}{x} \) yields \( \ln|x| \).
This relationship plays an essential role in our example. After the initial polynomial long division, we obtain a term \( \frac{5}{x+2} \). Its integral can be found using the formula for logarithmic integration, resulting in \( -5 \ln|x+2| \) in the final solution.
Understanding this integration process involves recognizing:

• The integrand's form and its capability of fitting the \( \frac{1}{x} \) template.
• Applying the correct integration rule to handle the expression appropriately.
• Including absolute values to ensure the expression remains valid for all potential inputs.

Mastering natural logarithm integration unlocks deeper analytical capability across a variety of calculus problems, particularly within rational function integrals.