Chapter 6: Problem 14
Find each integral by using the integral table on the inside back cover. $$ \int \frac{x}{(x+1)(x+2)} d x $$
Short Answer
Expert verified
\( \int \frac{x}{(x+1)(x+2)} dx = \ln\left( \frac{(x+2)^2}{x+1} \right) + C \)
Step by step solution
01
Identify the Formula
This integral involves a rational function, and it suggests the use of partial fraction decomposition. The integral resembles the format found in integral tables where the denominator is a product of linear factors.
02
Decompose into Partial Fractions
To use partial fractions, express the integrand \( \frac{x}{(x+1)(x+2)} \) as \( \frac{A}{x+1} + \frac{B}{x+2} \). Multiply through by the common denominator to get \( x = A(x+2) + B(x+1) \).
03
Determine the Constants A and B
Expand the equation from Step 2: \( x = Ax + 2A + Bx + B \) which simplifies to \( x = (A + B)x + (2A + B) \). Comparing coefficients, set \( A + B = 1 \) and \( 2A + B = 0 \).
04
Solve for A and B
Solve the system of equations: - From \( A + B = 1 \), we get \( B = 1 - A \).- Substitute into \( 2A + B = 0 \) to get \( 2A + (1 - A) = 0 \), which simplifies to \( A = -1 \), and thus \( B = 2 \).
05
Rewrite the Integral with Partial Fractions
Substituting back the values of A and B into the partial fractions, the integral becomes \( \int \left( \frac{-1}{x+1} + \frac{2}{x+2} \right) dx \).
06
Integrate Partial Fractions
Integrate each term separately: - \( \int \frac{-1}{x+1} dx = -\ln|x+1| + C_1 \)- \( \int \frac{2}{x+2} dx = 2\ln|x+2| + C_2 \)Combine the constants to give the final expression: \( -\ln|x+1| + 2\ln|x+2| + C \).
07
Simplify and State the Final Solution
Combine the two logarithmic terms using logarithmic properties: \[ 2\ln|x+2| - \ln|x+1| = \ln\left( \frac{(x+2)^2}{x+1} \right) \]The final expression for the integral is: \[ \ln\left( \frac{(x+2)^2}{x+1} \right) + C \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Fraction Decomposition
Partial fraction decomposition is a technique often used in calculus to break down complex rational functions into simpler fractions that are more accessible for integration. Imagine you have a fraction with a polynomial in the numerator and another in the denominator. The goal is to express this complex fraction as a sum of simpler fractions, which are easier to manage.
- Start by expressing your rational function as a sum of fractions with unknown constants.
- Each term in the denominator has its own fraction in the decomposition.
- For distinct linear factors in the denominator, assign a separate constant for each.
Integral Table
An integral table is a helpful tool for finding the antiderivatives of many common functions. They are especially useful in cases where a function matches a form listed in the table, making the integration process quicker and easier.
- Integral tables list many functions along with their corresponding integrals.
- They often cover various functions, such as exponential, trigonometric, and logarithmic forms.
- Make sure to adjust constants and coefficients to match the table accurately.
Rational Functions
Rational functions, a staple in calculus, are quotients of two polynomials. These functions are characterized by certain behaviors which make them interesting and sometimes challenging.
- They can have vertical asymptotes where the denominator equals zero.
- Their behavior at large values of \( x \) depends on the degrees of the numerator and denominator.
- Integration often requires special techniques like partial fraction decomposition.