Question

Use the technique of completing the square to evaluate the following integrals. $$ \int \frac{1}{x^{2}+2 x+1} d x $$

Short answer

The integral evaluates to \( -\frac{1}{x+1} + C \).

Step-by-step solution

Identify the Quadratic Expression

The given integral has the expression \( x^2 + 2x + 1 \) in the denominator. Notice that this is a quadratic expression.

Complete the Square

To complete the square, we rewrite the quadratic expression \( x^2 + 2x + 1 \). We recognize that it can be rewritten as \( (x + 1)^2 \) because \((x + 1)^2 = x^2 + 2x + 1\).

Rewrite the Integral

Substitute the completed square into the integral to get \( \int \frac{1}{(x+1)^2} \, dx \).

Integrate the Expression

The integral \( \int \frac{1}{(x+1)^2} \, dx \) can be evaluated using the formula for integrating \( \frac{1}{u^2} \), where \( u = x + 1 \). The integral becomes \( -\frac{1}{x+1} + C \), where \( C \) is the constant of integration.

Key concepts

Completing the Square

Completing the square is a technique used to manipulate quadratic expressions into a perfect square trinomial form. This makes them easier to work with, especially when integrating or solving equations. Let's break it down:

• A quadratic expression generally looks like this: \( ax^2 + bx + c \).
• To complete the square, you want to transform it into the form \((x + d)^2 + e\).
• For the expression \( x^2 + 2x + 1 \), first identify the coefficient of \( x \), which is 2. Divide it by 2, giving 1, and square it, which is still 1.

You can now rewrite \( x^2 + 2x + 1 \) as \((x+1)^2\).

Completing the square is useful because it simplifies the expression, making it easier to integrate or differentiate. This technique turns an expression that might look complex into something more familiar and manageable.

Integral Transformation

Integral transformation is a key step in solving integrals. It involves changing an integral into a simpler or more usable form. Essentially, you're rewriting the integral into something easier to evaluate. Here's how this helps:

• Once a quadratic expression is completed as a square, it often becomes a known function form.
• You substitute this form back into the integral, in effect transforming it.

In our exercise, rewriting \( \int \frac{1}{x^2 + 2x + 1} \, dx \) to \( \int \frac{1}{(x+1)^2} \, dx \) took advantage of completing the square.

Transformations like these simplify the integration process by bringing the integral into a form that has known solutions or rules for integration.

Definite Integrals

Definite integrals form the concept of finding the area under a curve within given limits. Though our exercise focused on indefinite integrals, understanding definite integrals is crucial for comprehension. Here's a short rundown of the concept:

• A definite integral is expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
• The main goal is to calculate the net area between \( f(x) \) and the x-axis from \( x = a \) to \( x = b \).

When evaluating definite integrals, you apply the fundamental theorem of calculus and calculate the difference between the antiderivative evaluated at \( b \) and \( a \).

Although definite integrals were not directly part of this exercise, knowing how to apply them with methods like completing the square is essential in advanced calculus.

Constant of Integration

Whenever you compute an indefinite integral, you need to include a constant of integration. This constant is vital because an antiderivative represents a family of functions that differ by a constant. Here's why it's important:

• An antiderivative of a function \( f(x) \) doesn't specify just one function, but rather infinitely many functions differing by a constant.
• The notation \( + C \) in the antiderivative \( -\frac{1}{x+1} + C \) covers this family of functions.

The constant \( C \) ensures the generality of the solution, meaning it accounts for all possible vertical shifts of the antiderivative graph. Ignoring this constant would lead to an incomplete solution for any indefinite integral problem.

Thus, always remember, while "+ C" might seem negligible, it preserves the mathematical accuracy and integrity of the solution.