Problem 44 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{x^{2}+3 x+2}{x\left(x^{2}+2 x+2\right)} d x$$

Step-by-Step Solution

Answer

Question: Evaluate the integral \(\int \frac{x^2 + 3x + 2}{x(x^2 + 2x + 2)}dx\). Answer: \(\int \frac{x^2 + 3x + 2}{x(x^2 + 2x + 2)}dx = \ln|x| + \arctan(x+1) + C\)

Step 1: Perform partial fraction decomposition

We have the integrand as: \(\frac{x^2 + 3x + 2}{x(x^2 + 2x + 2)}\). We will now perform partial fraction decomposition on this integrand. $$\frac{x^{2}+3x+2}{x\left(x^{2}+2x+2\right)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 2x + 2}$$ To find the values of coefficients A, B, and C, we multiply both sides by the common denominator \(x\left(x^{2}+2x+2\right)\): $$x^2 + 3x + 2 = A(x^2 + 2x + 2) + (Bx + C)(x)$$ Expanding the equation, we get: $$x^2 + 3x + 2 = Ax^2 + 2Ax + 2A + Bx^2 + Cx$$ Now, we group the terms of the same degree and compare the coefficients: $$x^2: \ 1 = A + B$$ $$x: \ 3 = 2A+C$$ $$const: \ 2 = 2A$$

Step 2: Solve for the values of the variables A, B, and C

Now we have a system of three linear equations: $$A + B = 1$$ $$2A + C = 3$$ $$2A = 2$$ From the third equation, we get the value of A immediately as \(A=1\). Substituting A's value into the first and second equation, we get: $$1 + B = 1 \implies B = 0$$ $$2(1) + C = 3 \implies C=1$$ Now that we have the values for A, B, and C, our original integrand can be expressed as: $$\frac{1}{x} + \frac{0x + 1}{x^2 + 2x + 2}$$

Step 3: Integrate the simpler fractions

Now we will integrate each term of the decomposed integrand separately: $$\int \frac{1}{x} dx = \ln|x| + C_1$$ For the second term, we can rewrite it as: $$\int \frac{1}{x^{2}+2 x+2} d x$$ We can complete the square in the denominator: $$\int \frac{1}{(x+1)^2 + 1}dx$$ This is a known form of an inverse tangent integral. The integral can be evaluated as: $$\int \frac{1}{(x+1)^2 + 1}dx = \arctan (x+1) + C_2$$

Step 4: Combine the results

Now we can combine the results of the two integrals: $$\int \frac{x^{2}+3 x+2}{x\left(x^{2}+2 x+2\right)} d x = \ln|x| + \arctan(x+1) + C$$ Where C is the constant of integration, given by \(C = C_1 + C_2\).

Key Concepts

Integration Techniques

Integration techniques are diverse methods used to evaluate complex integrals, breaking them down into simpler forms. In this problem, we utilize **Partial Fraction Decomposition**, a strategy to simplify rational expressions. This technique involves expressing a complicated rational function as a sum of simpler fractions, which are easier to integrate.

**Partial Fraction Decomposition** is particularly beneficial when the denominator can be factored. Here, the integrand \(\frac{x^2 + 3x + 2}{x(x^2 + 2x + 2)}\) is split into two simpler fractions: \(\frac{A}{x} + \frac{Bx + C}{x^2 + 2x + 2}\). By solving for the constants \(A, B,\) and \(C\), we transform the integrand into an easy-to-integrate form.

Steps involve:

• Writing the rational function in terms of partial fractions.
• Equating coefficients from both sides.
• Finding constants through a system of equations.
• Integrating each simpler fraction separately.

Inverse Trigonometric Functions

Inverse trigonometric functions allow the computation of integrals involving expressions derived from trigonometric identities. In our problem, we dealt with such a function by rewriting the term \(\int \frac{1}{x^2 + 2x + 2} dx\) using the inverse tangent formula.

To proceed, we first complete the square in the denominator, transforming \(x^2 + 2x + 2\) into \((x+1)^2 + 1\). Recognizing it as a standard integral of the form \(\int \frac{1}{(u)^2 + a^2} du\), we apply the formula:

• \(\int \frac{1}{(u)^2 + a^2} du = \frac{1}{a} \arctan \left(\frac{u}{a}\right) + C\)

This results in:

• \(\int \frac{1}{(x+1)^2 + 1} dx = \arctan(x+1) + C_2\)

Such transformations simplify integration and emphasize the power of inverse trigonometric functions in calculus.

Calculus

Calculus, the mathematical study of change, includes techniques for evaluation, like differentiation and integration. The problem given is an integral, forming a critical part of calculus, where we seek to find the area under a curve or the accumulated sum of infinitesimal parts.

Integration, central in calculus, involves finding a function (antiderivative) whose derivative is the integrand. Complex integrals, such as \(\int \frac{x^2 + 3x + 2}{x(x^2 + 2x + 2)} dx\), are tackled by breaking them into simpler forms through methods like partial fraction decomposition, leading to straightforward integration.

In solving integrals, it's important to:

• Understand various techniques, such as substitution, integration by parts, and partial fractions.
• Use algebraic manipulations to simplify expressions.
• Apply inverse trigonometric functions when suitable.

This exercise highlights calculus' essence in transforming a complex expression into understandable parts, demonstrating integration’s fundamental role within mathematical analysis.