Problem 51 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-54, perform the indicated integrations. \(\int \frac{x+1}{9 x^{2}+18 x+10} d x\)

Step-by-Step Solution

Answer

\( \int \frac{x+1}{9x^2 + 18x + 10} \, dx = \frac{1}{3} \tan^{-1}(3(x+1)) + \frac{1}{18} \ln|9(x+1)^2+1| + C \).

Step 1: Identify the form of the integrand

We have the integral \( \int \frac{x+1}{9x^2 + 18x + 10} \, dx \). The denominator is a quadratic expression that might be factorable or lend itself to partial fraction decomposition.

Step 2: Attempt to factor the quadratic

The quadratic expression in the denominator is \( 9x^2 + 18x + 10 \). We check if it can be factored, possibly using trial methods or by finding roots if possible.

Step 3: Check if the quadratic can be simplified or factored

Unfortunately, \( 9x^2 + 18x + 10 \) does not factor easily, so we will check if completing the square is feasible or consider a substitution method.

Step 4: Complete the square for the denominator

Write \( 9x^2 + 18x + 10 \) as \( 9(x^2 + 2x) + 10 \). Completing the square inside gives \( 9((x+1)^2 - 1) + 10 \), which simplifies to \( 9(x+1)^2 + 1 \).

Step 5: Use substitution to simplify the integral

Let \( u = x + 1 \) so that \( du = dx \) and \( x = u - 1 \). Substitute into the integral to get \( \int \frac{u}{9u^2 + 1} du \).

Step 6: Split the integral into two parts

Re-write the integral as \( \int \frac{1}{9u^2 + 1} \, du + \int \frac{u}{9u^2 + 1} \, du \).

Step 7: Solve each integral separately

The first integral \( \int \frac{1}{9u^2 + 1} \, du \) can be solved by recognizing it as an arctan form: \( \frac{1}{3} \int \frac{1}{u^2 + \left(\frac{1}{3}\right)^2} \, du = \frac{1}{3} \tan^{-1}(3u) + C_1 \). The second integral \( \int \frac{u}{9u^2 + 1} \, du \) can be solved by substitution and results in \( \frac{1}{18} \ln|9u^2+1| + C_2 \).

Step 8: Combine the solutions and back-substitute

Combine the terms: \( \frac{1}{3} \tan^{-1}(3(x+1)) + \frac{1}{18} \ln|9(x+1)^2+1| + C \).

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a vital integration technique used when dealing with rational functions. It transforms complex fractions into simpler parts, making integration easier.
The process involves expressing a fraction like \( \frac{P(x)}{Q(x)} \) as a sum of simpler fractions, where \( Q(x) \) is a polynomial that can be factorized.

• If the denominator is a product of distinct linear factors, each term takes the form \( \frac{A}{x-a} \).
• If there are repeated linear factors, terms take the form \( \frac{A}{x-a} + \frac{B}{(x-a)^2} + \dots \).
• Quadratic factors that can't be easily factorized are handled using another method, like completing the square or trigonometric substitution.

In the given exercise, partial fraction decomposition was initially considered, but the denominator \( 9x^2 + 18x + 10 \) wasn't suitable due to its inability to be factorized into real simple fractions.
This technique helps break down complex problems, simplifying calculations.

Completing the Square

Completing the square is a technique used to simplify integration, especially when dealing with quadratic polynomials. Here’s how it works:
First, express the quadratic in the form \( ax^2 + bx + c \). The aim is to rewrite it as \( a(x-h)^2 + k \).

• Take the coefficient \( b \) of \( x \), divide by 2, and square it.
• Add and subtract this square inside the polynomial to form a perfect square trinomial.

In our exercise, the quadratic \( 9x^2 + 18x + 10 \) was rewritten as \( 9(x+1)^2 + 1 \).
This transformation facilitated other steps, because it made the function easy to handle with trigonometric substitution.

Trigonometric Substitution

Trigonometric substitution is a clever integration technique where trigonometric identities are used to simplify integrals involving square roots and quadratic forms.
This method is often applied to expressions of the forms \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), or \( \sqrt{x^2 - a^2} \).

• In cases like \( \sqrt{a^2 + x^2} \), a substitution such as \( x = a \tan(\theta) \) is used.
• These substitutions transform the integrand into a trigonometric form that can be more manageable.

In the step-by-step solution, recognizing the form \( \int \frac{1}{9u^2 + 1} \, du \) helped in using the arctan integration formula to solve it.
This conversion made solving the integral possible using simpler techniques related to inverse trigonometric functions.

Definite and Indefinite Integrals

In calculus, integration comes in two types: definite and indefinite integrals. Both are central to solving a wide array of problems.
Indefinite Integrals: These represent antiderivatives and include the constant of integration \( C \).

• General solution without specific limits.
• Example: \( \int x^2 \, dx = \frac{x^3}{3} + C \).

Definite Integrals: These have specific limits and calculate the area under a curve between two points.

• No constant \( C \), and solved using limits.
• Example: \( \int_{0}^{1} x^2 \, dx = \frac{1}{3} \).

In the exercise, indefinite integrals are considered as the solution includes a general antiderivative with \( C \).
Recognizing the difference aids in understanding the purpose and application of each integral type in calculus.