Problem 14 - Calculus for Biology and Medicine

Question

Use the Table of Integrals to compute each integral after manipulating the integrand in a suitable way. $$ \int \frac{x^{2}}{x^{2}+4 x+1} d x $$

Step-by-Step Solution

Answer

The integral is \( (x + 2) - 2\ln|(x^2 + 4x + 1)| + C \).

Step 1: Complete the Square

First, observe the denominator \(x^2 + 4x + 1\). We complete the square to rewrite it in a more manageable form. Completing the square involves expressing \(x^2 + 4x + 1\) as \((x+b)^2 - c\).The expression \(x^2 + 4x + 1\) can be rewritten as \((x+2)^2 - 3\), because: \[(x+2)^2 = x^2 + 4x + 4\]So: \[x^2 + 4x + 1 = (x+2)^2 - 3\]

Step 2: Substitute for Simplicity

Let \(u = x + 2\), then \(du = dx\). Adjust the integral bounds accordingly, this changes \(x^2 + 4x + 1\) to \(u^2 - 3\). Hence:\[\int \frac{x^2}{x^2+4x+1} \, dx = \int \frac{(u-2)^2}{u^2 - 3} \, du\]

Step 3: Expand the New Integrand

Expand \((u-2)^2\) to make substitution easier:\[(u-2)^2 = u^2 - 4u + 4\]This makes the integrand:\[\frac{(u^2 - 4u + 4)}{u^2 - 3} = 1 - \frac{4u - 4}{u^2 - 3}\]

Step 4: Simplify the Fraction

Now split the fraction:\[\int 1 \, du - \int \frac{4(u-1)}{u^2-3} \, du = \int 1 \, du - 4 \int \frac{u-1}{u^2-3} \, du\]

Step 5: Integrate the Separable Parts

The integral of 1 with respect to \(u\) is simply \(u\):\[\int 1 \, du = u\]The second integral can be computed using substitution: let \(v = u^2 - 3\), then \(dv = 2u \, du\), or \(du = \frac{dv}{2u}\).This transforms the second integral:\[\int \frac{u-1}{u^2-3} \, du = \int \frac{1}{2} \left(\frac{2u - 2}{u^2 - 3}\right) \, du\]Thus we substitute and find:\[- \int \frac{dv}{v} = -\ln|v| = -\ln|u^2 - 3|\]Combining, we have:\[ u - 2\ln|u^2 - 3| + C\]

Step 6: Revert the Substitution

Replace \(u\) with \(x+2\) to return to the original variable:\[(x+2) - 2\ln|(x+2)^2 - 3| + C\]

Step 7: Final Answer Verification

Combine all parts to express the integral in terms of x:\[(x + 2) - 2\ln|(x^2 + 4x + 1)| + C\]This is the integral of the given function, simplified and expressed in terms of \(x\).

Key Concepts

Completing the Square

Completing the square is a useful algebraic technique for simplifying quadratic expressions. It helps by converting a quadratic polynomial into a simpler form that is easier to integrate. Here, our expression is given by the denominator of the integrand, \(x^2 + 4x + 1\). We aim to rewrite this as \((x+b)^2 - c\).
To complete the square for \(x^2 + 4x + 1\), we first find the perfect square trinomial. Begin by half of the coefficient of \(x\), which is \(4\), and squaring it: \((2)^2 = 4\). This results in the expression \((x+2)^2\), which expands to \(x^2 + 4x + 4\).
To form our original denominator inside the integral, we find:

• Original: \(x^2 + 4x + 1\)
• Formed Square: \((x+2)^2 - 3\)

Thus, the quadratic expression is now rewritten as \((x+2)^2 - 3\), making integration easier.

Substitution Method

The substitution method is a vital technique in integration for simplifying expressions. Here, we make the integral more manageable by changing variables. In our problem, we substitute \(u = x + 2\), which simplifies handling the squared term in our completed square expression \((x+2)^2 - 3\).
This substitution converts the derivative part \(du = dx\), maintaining the relationship with respect to \(x\). The integral then takes on a new form:

• Converts \(x^2 + 4x + 1\) to \(u^2 - 3\)
• Results in integrating \( \int \frac{(u-2)^2}{u^2 - 3} \, du \)

By substitution, we ease the process of integration while ensuring all parts align with the new variable, \(u\). This change simplifies the otherwise complex interaction between numerator and denominator.

Integration Techniques

Integration techniques involve various strategies to solve integrals that aren't immediately straightforward. After substituting, our task is to integrate the transformed expression.
First, we expand \((u-2)^2 = u^2 - 4u + 4\), leading to the decomposition of our integrand. It splits into re-integratable structures:

• \( \int \frac{(u^2 - 4u + 4)}{u^2 - 3} \, du \) becomes \( \int 1 \, du - 4 \int \frac{u - 1}{u^2 - 3} \, du \)

The integration splits into simpler parts that can be handled individually. Integrating \(1\) with respect to \(u\) straightforwardly results in \(u\). For \(\frac{u-1}{u^2-3}\), further substitution helps, simplifying complex expressions into manageable logarithmic integrations, revealing the form \(- \ln|u^2 - 3|\).
These strategies underline the importance of breaking down the integral into parts that are easier to handle, enabling step-by-step integration.

Rational Functions

Rational functions are fractions where the numerator and the denominator are polynomials. They often appear in calculus problems requiring advanced integration techniques like substitution or partial fraction decomposition.
Here, our rational function \( \frac{x^2}{x^2 + 4x + 1} \) lacks the apparent ease of simple integration due to its complexity. By completing the square in the denominator, we uncover a path to simplify and transform the problem for easier integration.
Integrating rational functions like this often involves steps: transforming the expression through algebraic manipulation, substitution to simplify, and applying specific integration strategies tailored to the transformed function. Each step reduces the difficulty inherent to rational functions, improving the chance of a successful integral calculation.