Problem 67 - Calculus for Biology and Medicine

Question

Evaluate each integral. $$ \int \frac{1}{x^{2}\left(x^{2}+1\right)} d x $$

Step-by-Step Solution

Answer

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

Step 1: Factorize the Denominator

The integrand is \( \frac{1}{x^2 (x^2 + 1)} \). Notice that the denominator \( x^2(x^2 + 1) \) involves the product of \( x^2 \) and \( x^2 + 1 \). We'll perform partial fraction decomposition.

Step 2: Set Up Partial Fractions

Express \( \frac{1}{x^2(x^2 + 1)} \) as a sum of simpler fractions: \[ \frac{1}{x^2(x^2 + 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 1} \].

Step 3: Simplify and Solve for Coefficients

Multiply through by the denominator \( x^2(x^2 + 1) \) to clear fractions: \[ 1 = A(x)(x^2 + 1) + B(x^2 + 1) + (Cx + D)x^2 \]. Expand and collect terms in descending powers of \( x \): \[ 1 = Ax^3 + Ax + Bx^2 + B + Cx^3 + Dx^2 \]. Rearrange: \[ 1 = (A + C)x^3 + (B + D)x^2 + (A)x + B \].

Step 4: Equate Coefficients

Equate coefficients of corresponding powers of \( x \) from both sides: - For \( x^3: A + C = 0 \)- For \( x^2: B + D = 0 \)- For \( x: A = 0 \)- Constant term: B = 1 From these, \( A = 0 \), \( B = 1 \), \( D = -1 \), and \( C = 0 \).

Step 5: Rewrite as Partial Fractions

Substituting back the values of the coefficients, the partial fraction becomes: \[ \frac{1}{x^2(x^2 + 1)} = \frac{1}{x^2} - \frac{1}{x^2 + 1} \].

Step 6: Integrate Each Term Separately

Now integrate term by term: \[ \int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = -x^{-1} = -\frac{1}{x} + C_1 \], \[ \int \frac{1}{x^2 + 1} \, dx = \tan^{-1}(x) + C_2 \].

Step 7: Combine the Integrals

Add the results of the integrals together: \[ -\frac{1}{x} - \tan^{-1}(x) + C \] where \( C = C_1 + C_2 \) is the constant of integration.

Key Concepts

Integral Calculus

Integral calculus is a fundamental branch of calculus that focuses on the concept of integrals. Integrals are used to find areas, volumes, and central points. They also help solve differential equations. The two main types of integrals are definite and indefinite integrals. In this context, we are dealing with an indefinite integral.

Indefinite integrals, also known as antiderivatives, reverse the process of differentiation. By doing this, they allow us to reconstruct a function given its derivative. Integral calculus is particularly powerful in solving real-life problems. It helps in calculating displacement given velocity, or finding the accumulated quantity over time.

When we evaluate an indefinite integral, the result includes a constant of integration. This constant is crucial because an infinite number of functions can have the same derivative. Thus, the constant represents all possible vertical shifts of the antiderivative.

Integration Techniques

Integration techniques are strategies used to simplify and solve complex integrals. In the problem given, we used partial fraction decomposition. This technique is particularly handy for integrating rational functions. When a function is expressed as a single fraction, partial fraction decomposition breaks it down into simpler fractions. This makes it easier to integrate each part individually.

Partial fraction decomposition involves several steps:

• First, factorize the denominator.
• Next, set up an equation expressing the function as a sum of simpler fractions.
• Solve for the unknown coefficients by clearing fractions and equating coefficients of corresponding powers.

Once decomposed, we can integrate each term separately, often using basic integral formulas. Partial fraction decomposition is just one of many integration techniques. Others include substitution, integration by parts, and trigonometric substitution. Each technique has its place and utility, depending on the form of the integrand.

Definite and Indefinite Integrals

In integral calculus, it's crucial to distinguish between definite and indefinite integrals.

**Indefinite Integrals**
- Indefinite integrals provide a family of functions that differentiate to the integrand. They include a constant of integration, represented as 'C".
- The notation for indefinite integrals is \(\int f(x) \, dx = F(x) + C\). This implies that the derivative of \(F(x)\) is \(f(x)\).

**Definite Integrals**
- Definite integrals, on the other hand, have limits. They yield a numerical value representing the net area under a curve between two points.
- Denoted as \(\int_{a}^{b} f(x) \, dx\), it calculates the accumulation of the quantity over a specific interval.
- The result of a definite integral does not include a constant of integration, as it gives a single number, not a function.
Understanding these differences is important for solving problems correctly. While our given exercise deals with an indefinite integral, recognizing how definite integrals differ helps in broader applications of calculus.