Problem 57 - Calculus for Biology and Medicine

Question

Use either substitution or integration by parts to evaluate each integral. $$ \int \frac{x}{x+3} d x $$

Step-by-Step Solution

Answer

\( \int \frac{x}{x+3} \, dx = x + 3 - 3 \ln |x + 3| + C \).

Step 1: Choose a Method

In order to solve the integral \( \int \frac{x}{x+3} \, dx \), we first determine which method to use. Substitution may be useful here as the derivative of \( x+3 \) is 1, which closely matches part of the integrand \( x \).

Step 2: Setup the Substitution

Let's use substitution and set \( u = x + 3 \). Then, the differential \( du = dx \) and \( x = u - 3 \). Substitute these expressions into the integral to obtain:\[ \int \frac{x}{x+3} \, dx = \int \frac{u - 3}{u} \, du \].

Step 3: Simplify the Integral

Simplify the expression by separating it:\[ \int \frac{u - 3}{u} \, du = \int \left( 1 - \frac{3}{u} \right) \, du = \int 1 \, du - \int \frac{3}{u} \, du \].

Step 4: Integrate Each Term

Evaluate each integral separately:- The integral \( \int 1 \, du \) evaluates to \( u \).- The integral \( \int \frac{3}{u} \, du \) evaluates to \( 3 \ln |u| \).Thus, the overall expression is \( u - 3 \ln |u| \).

Step 5: Substitute Back to Original Variable

Substitute \( u = x + 3 \) back into the expression:\[ u - 3 \ln |u| = (x + 3) - 3 \ln |x + 3| \].

Step 6: Write the Final Answer

Add the constant of integration \( C \) because this is an indefinite integral:\[ \int \frac{x}{x+3} \, dx = x + 3 - 3 \ln |x + 3| + C \].

Key Concepts

Integration by Parts

Integration by parts is a fundamental technique used in calculus to integrate products of functions. It is analogous to the product rule in differentiation. Rather than solving the integral directly, it allows us to transform the integral of a product into a different integral that is hopefully easier to evaluate.

The formula for integration by parts is given by:

• \[ \int u \, dv = uv - \int v \, du \]

Here, we identify parts of the integrand as \( u \) and \( dv \), then derive \( du \) and \( v \) to use in the formula. This technique is particularly useful when an integral contains a product of algebraic, exponential, logarithmic, or trigonometric functions.

In this context, the exercise was solved using substitution, but understanding integration by parts is essential, as it could be applied in similar integrals, especially when substitution is not straightforward.

Indefinite Integral

An indefinite integral refers to the antiderivative of a function. It represents a family of functions whose derivative is the given function. The notation for the indefinite integral of a function \( f(x) \) is:

• \[ \int f(x) \, dx \]

The result is an antiderivative \( F(x) \) plus a constant of integration, denoted by \( C \). This constant is necessary because differentiating removes any constant, and so, integrating adds a constant back to account for it.

In the original problem, after simplifying and integrating each term, the result was combined to give the solution, recognizing it as an indefinite integral by adding the constant \( C \). This highlights that the integral’s specific solution depends on initial conditions, allowing for all possible variations of the antiderivative.

Integral Calculus

Integral calculus is the branch of calculus that concerns accumulation of quantities, such as areas under curves and other applications involving summation. It is one of the main components of calculus, the other being differential calculus.

The primary goals of integral calculus are calculating area, volume, and solving differential equations through integration. Two major concepts in integral calculus are:

• Definite integrals, which involve integration over a specific interval and result in a number.
• Indefinite integrals, like the one we’re examining, that involve finding a general function whose derivative gives the original function.

In our exercise, we examined an indefinite integral, demonstrating how calculus techniques simplify complex functions by accumulating their values over intervals. This powerful mathematical tool forms the basis for solving real-world problems involving changing quantities or continuously varying rates.