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Mobile App Chapter 17: Problem 4
By expressing the following in partial fractions evaluate the given integral. Remember to select the correct form for the partial fractions. $$ \int \frac{2 x}{(x-1)^{2}(x+1)} \mathrm{d} x $$
Short Answer
Expert verified
The integral of the function \( \frac{2x}{(x-1)^2 (x+1)} \) is:
$$
-\frac{1}{x-1} - \frac{1}{2} \ln |x+1| + C
$$
Step by step solution
01
Decompose the integrand into partial fractions
To decompose the given integrand into partial fractions, we need to find two constants A and B such that:
$$
\frac{2x}{(x-1)^2 (x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1}
$$
Now, we will clear the denominators by multiplying both sides by \((x-1)^2 (x+1)\):
$$
2x = A(x-1)(x+1) + B(x+1) + C(x-1)^2
$$
This is the equation we need to solve for A, B, and C.
02
Solve for A, B, and C
To find A, B, and C, we can substitute values for x that will eliminate the other terms:
For A, let x = 1:
$$
2 = A(0)
$$
We can see that A is not necessary as the equation results in 0.
For B, let x = 1:
$$
2 = B(2) \Rightarrow B = 1
$$
For C, let x = -1:
$$
-2 = C(-2)^2 \Rightarrow C = -\frac{1}{2}
$$
Now we have:
$$
\frac{2x}{(x-1)^2 (x+1)} = \frac{1}{(x-1)^2} - \frac{1}{2(x+1)}
$$
03
Integrate each term separately
Now, we can integrate each term separately:
$$
\int \frac{2x}{(x-1)^2 (x+1)} \,\mathrm{d}x = \int \frac{1}{(x-1)^2} \,\mathrm{d}x - \frac{1}{2} \int \frac{1}{x+1} \,\mathrm{d}x
$$
The first integral is a standard formula: \(\int \frac{1}{x^2} \,\mathrm{d}x = -\frac{1}{x} + C_1\)
$$
\int \frac{1}{(x-1)^2} \,\mathrm{d}x = -\frac{1}{x-1} + C_1
$$
The second integral is also a standard formula: \(\int \frac{1}{x} \,\mathrm{d}x = \ln |x| + C_2\)
$$
-\frac{1}{2} \int \frac{1}{x+1} \,\mathrm{d}x = -\frac{1}{2} \ln |x+1| + C_2
$$
04
Combine the two integrals
Now, we combine the two integrals to get the final result:
$$
\int \frac{2x}{(x-1)^2 (x+1)} \,\mathrm{d}x = -\frac{1}{x-1} - \frac{1}{2} \ln |x+1| + C
$$
So, the integral of the given function is:
$$
-\frac{1}{x-1} - \frac{1}{2} \ln |x+1| + C
$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integral Calculus
Integral calculus is a branch of calculus focused on the process of finding the integral of functions. The integral, symbolized by the integral sign \( \int \), represents the accumulation of quantities, such as areas under a curve, total distance covered, or mass of an object with varying density. There are two main types of integrals: indefinite and definite. An indefinite integral involves finding a function (the antiderivative) that, when differentiated, yields the original function. It represents a family of functions rather than a single value and includes a constant of integration, often denoted by \( C \).
Definite integrals, on the other hand, are concerned with the calculation of net quantities between specific bounds, producing a distinct numerical value rather than a function. Integral calculus is key in mathematics as it allows us to solve problems that involve accumulation and is often complementary to differential calculus, providing solutions to the area, volume, and other physical concepts.
Definite integrals, on the other hand, are concerned with the calculation of net quantities between specific bounds, producing a distinct numerical value rather than a function. Integral calculus is key in mathematics as it allows us to solve problems that involve accumulation and is often complementary to differential calculus, providing solutions to the area, volume, and other physical concepts.
Algebraic Fraction Decomposition
Understanding Partial Fractions
Algebraic fraction decomposition, also known as partial fraction decomposition, is an essential technique in integral calculus when dealing with rational functions (quotients of polynomials). The goal is to break down a complex fraction into simpler, more manageable 'partial fractions' that are easier to integrate. This process involves expressing the original fraction as a sum of fractions with simpler denominators. The process starts by determining a template for the partial fractions based on the factors of the original denominator and then solving for constants in the numerators that will satisfy the identity. This technique is especially useful when the degree of the numerator is less than the degree of the denominator, allowing for an eventual integration of standard forms.Indefinite Integral
An indefinite integral represents a family of functions that reverses the process of differentiation---that is, the original function is recovered when the derivative of the indefinite integral is taken. The general notation \( \int f(x) \, \mathrm{d}x \) is used to denote the indefinite integral of \( f(x) \), where \( f(x) \) is called the integrand. Solving an indefinite integral, or integrating a function, often involves manipulating the function into a form for which the antiderivative is known. The integration process comes with a unique constant of integration, simply denoted as \( C \) since the derivative of any constant is zero, preserving the correctness of the antiderivative. This concept is critical in the practice of integral calculus and appears in a multitude of applications across engineering, physics, economics, and beyond.
