Problem 51 - Calculus for Biology and Medicine

Question

$$ \text { In Problems , evaluate each integral. } $$ $$ \int \frac{1}{x^{2}\left(x^{2}+1\right)} d x $$

Step-by-Step Solution

Answer

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

Step 1: Decompose the integrand

To integrate \( \frac{1}{x^2 (x^2 + 1)} \), first decompose it using partial fraction decomposition. We assume a form: \( \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 1} \). Set up the equation for the numerator: \( 1 = A x (x^2 + 1) + B (x^2 + 1) + (Cx + D)x^2 \).

Step 2: Simplify and equate coefficients

Expand and simplify the expression: \( 1 = Ax^3 + Ax + Bx^2 + B + Cx^3 + Dx^2 \). Combine like terms: \( 1 = (A + C)x^3 + (B + D)x^2 + Ax + B \). Equating coefficients from both sides, we get: \( A + C = 0 \), \( B + D = 0 \), \( A = 0 \), \( B = 1 \).

Step 3: Solve the system of equations

From the equations, we determine: \( A = 0 \), \( B = 1 \). Substituting \( A = 0 \) into \( A + C = 0 \) gives \( C = 0 \). Substituting \( B = 1 \) into \( B + D = 0 \) gives \( D = -1 \). The decomposition is thus: \( \frac{1}{x^2} - \frac{1}{x^2 + 1} \).

Step 4: Integrate the decomposed functions

Integrate each term separately: For \( \int \frac{1}{x^2} \, dx \), we get \( -\frac{1}{x} + C_1 \). For \( \int \frac{1}{x^2 + 1} \, dx \), we get \( -\tan^{-1}(x) + C_2 \).

Step 5: Combine the integrals

Combine the results of the integrals: The result is \( -\frac{1}{x} - \tan^{-1}(x) + C \), where \( C = C_1 + C_2 \) is the constant of integration.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a valuable technique in Integral Calculus used to simplify the integration of rational functions. This method breaks down complex fractions into simpler parts, making them easier to integrate individually. The fundamental idea is to express the rational function as a sum of simpler fractions. Let's say you have a fraction of the form \(\frac{P(x)}{Q(x)}\), where \(Q(x)\) must be factored into linear or irreducible quadratic factors. For example, in the problem \(\int \frac{1}{x^2(x^2 + 1)} \, dx\), we decomposed it as:

• \(\frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 1}\)

This involves finding constants \(A, B, C,\) and \(D\) by setting up an equation of equality with the original fraction's numerator and solving the resulting system of equations from matching the coefficients. This step is crucial because once the integrand is simplified into these fractional components, each piece can be integrated separately, leveraging simpler integral rules like power functions or basic trigonometric integrals.

Integrals of Rational Functions

Integrating rational functions is often challenging due to their complex structure. Rational functions are the ratio of two polynomials, for instance, \(\frac{1}{x^2(x^2 + 1)}\), which was dealt with in the original exercise. The process usually begins with partial fraction decomposition, which we previously described. This helps break down the function into manageable parts.Once decomposed, each fraction can often be integrated straightforwardly:

• The integral of \(\frac{B}{x^2}\) simplifies to \(-\frac{1}{x}\) since it follows the power rule for integrals.
• The term \(\frac{1}{x^2 + 1}\) is a standard integral that results in \(-\tan^{-1}(x)\), leveraging the inverse tangent function's differentiation property.

To finish, sum these individual integrals, along with a constant of integration \(C\), to form the complete integral solution.

Trigonometric Substitution

Trigonometric substitution is a technique used to solve integrals involving polynomial expressions that suggest a trigonometric identity. This technique is particularly useful when encountering forms like \(a^2 - x^2\), \(x^2 + a^2\), or \(x^2 - a^2\) within an integrand. While it wasn't directly utilized in the given exercise, understanding where it fits can be handy. For instance, with terms like \(x^2 + 1\), we might consider using trigonometric identities. The typical approach with a term like this involves setting \(x = a \tan(\theta)\), transforming the integral into one you can solve using trigonometric identities and derivatives. Trigonometric substitution converts a difficult algebraic integral into a simpler trigonometric one, which can often be resolved using direct substitution or inversion formulas. This method provides a powerful tool in tackling integrals that do not easily break down using partial fractions alone, especially when confronting square roots within the integrand.