Problem 82 - Calculus and its applications

Question

Evaluate. Use the technique of Example \(9 .\) $$ \int \frac{x+3}{x-2} d x \quad(\text { Hint: } u=x-2 .) $$

Step-by-Step Solution

Answer

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

Step 1 - Identify the Substitution

The problem suggests using the substitution technique where we define a new variable to simplify the integral expression. Given the hint, let us set the substitution as \( u = x - 2 \). This substitution simplifies the expression under the integral.

Step 2 - Differentiate the Substitution

To make the substitution work, we need to express \( dx \) in terms of \( du \). Differentiate the substitution \( u = x - 2 \) to get \( du = dx \).

Step 3 - Express x in Terms of u

Since \( u = x - 2 \), we can write \( x = u + 2 \). This helps us to replace \( x \) in the integral with expressions involving \( u \).

Step 4 - Rewrite the Integral

Substitute \( x = u + 2 \) and \( dx = du \) into the original integral to get: \[ \int \frac{u + 2 + 3}{u} \, du = \int \frac{u + 5}{u} \, du. \] Simplify this to: \[ \int \left( 1 + \frac{5}{u} \right) \, du. \]

Step 5 - Separate and Integrate

Divide the integral into two separate integrals: \[ \int 1 \, du + \int \frac{5}{u} \, du. \] The first integral, \( \int 1 \, du \), simplifies to \( u \), and the second integral, \( \int \frac{5}{u} \, du \), simplifies to \( 5 \ln|u| \). So the integral becomes: \[ u + 5 \ln|u| + C. \]

Step 6 - Substitute Back in Terms of x

Replace \( u \) with \( x - 2 \) to revert to the original variable: \[ (x - 2) + 5 \ln|x - 2| + C. \] This is the antiderivative in terms of \( x \).

Key Concepts

Substitution Method

The substitution method is a technique in calculus used to simplify integrals by changing the variables. When dealing with complex integral expressions, substituting with a new variable makes the equation easier to handle. For this particular exercise, the integral \( \int \frac{x+3}{x-2} \, dx \) was simplified using the hint provided with the problem: let \( u = x - 2 \).
By substituting \( u \) for \( x - 2 \), we can differentiate this equation and find that \( du = dx \). This allows us to rewrite both the integral and its differential components. The key steps here involve:

• Identifying an appropriate substitution often provided in the hint.
• Expressing the differential \( dx \) in terms of \( du \).
• Rewriting the entire integral in terms of \( u \), which simplifies the process of integration.

By changing the variable, the process of integrating becomes straightforward, which would otherwise be complicated by the algebraic form of the expression.

Definite Integrals

A definite integral computes the accumulation of quantities, like area under a curve, between specific limits. For this exercise, the use of a definite integral isn't explicitly mentioned, but understanding it is crucial as it often follows after finding the antiderivative.Definite integrals provide a way to find total quantities which are represented by algebraic expressions in calculus. When converting a problem to definite integrals:

• You replace the function with its antiderivative.
• You then apply the limits of integration, usually represented as \( \int_{a}^{b}( f(x) \, dx ) \), evaluating the antiderivative at points \( a \) and \( b \).
• The result is given by the difference between the upper and lower values at the boundaries of \( a \) and \( b \).

The fundamental theorem of calculus connects the derivative and integral, which simplifies the transition from finding antiderivatives to calculating definite integrals.

Antiderivative

Antiderivatives are functions that reverse differentiation, recovering the original function before it was derived. In the given exercise, after performing substitution and integration, finding the antiderivative means expressing the integral in terms of the original variable \( x \).In our resolved exercise:

• We converted \( u + 5 \ln|u| + C \) back into terms of \( x \) by substituting \( u = x - 2 \).
• This results in the solution \((x - 2) + 5 \ln|x - 2| + C\).
• The constant \( C \) represents an unknown constant of integration, as the antiderivative provides a family of functions differing by a constant.

Antiderivatives form the basis for solving many calculus problems by allowing one to determine a function when its rate of change is known. The process often involves finding integrals, differentiating strategically, and including the constant of integration.