Problem 26 - Calculus Single Variable

Question

Use the method of partial fractions to decompose the integrand. Then evaluate the given integral. \(\int \frac{2 x^{2}}{(x+1)^{2}\left(x^{2}+1\right)} d x\)

Step-by-Step Solution

Answer

The integral evaluates to \(-\ln|x+1| - \frac{1}{x+1} + \frac{1}{2} \ln|x^2+1| + C\).

Step 1: Setup the partial fraction decomposition

We begin by expressing \(\frac{2x^2}{(x+1)^2(x^2+1)}\) as a sum of simpler fractions. The form of the partial fraction decomposition for the expression is:\[\frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{Cx + D}{x^2 + 1}\]Our goal is to find the constants \(A, B, C,\) and \(D\) that satisfy this equation.

Step 2: Clear the denominators

Multiply both sides of the equation by the denominator \((x+1)^2(x^2+1)\) to clear the fractions:\[2x^2 = A(x+1)(x^2+1) + B(x^2+1) + (Cx+D)(x+1)^2\]Expand each term on the right-hand side of the equation.

Step 3: Expand and collect like terms

Expand each term on the right-hand side:- \(A(x+1)(x^2+1) = A(x^3 + x^2 + x + 1) = Ax^3 + Ax^2 + Ax + A\)- \(B(x^2+1) = Bx^2 + B\)- \((Cx+D)(x+1)^2 = (Cx+D)(x^2 + 2x + 1) = Cx^3 + 2Cx^2 + Cx + Dx^2 + 2Dx + D\)Combine these to get:\[Ax^3 + (A + B + 2C + D)x^2 + (A + 2D + C)x + (A + B + D)\]

Step 4: Equate coefficients

Compare coefficients from both sides of the equation to create a system of equations:- Coefficient of \(x^3\): \(A + C = 0\)- Coefficient of \(x^2\): \(A + B + 2C + D = 2\)- Coefficient of \(x\): \(A + 2D + C = 0\)- Constant term: \(A + B + D = 0\)

Step 5: Solve the system of equations

Solve the system of equations:1. From \(A + C = 0\), we have \(C = -A\).2. From \(A + 2D + C = 0\), substitute \(C = -A\) to get \(A + 2D - A = 0 \Rightarrow 2D = 0 \Rightarrow D = 0\).3. Substitute \(C = -A\) and \(D = 0\) into \(A + B + 2C + D = 2\) to get \(A + B - 2A = 2 \Rightarrow -A + B = 2\).4. Substitute \(D = 0\) into \(A + B + D = 0\) to get \(A + B = 0\).5. Solve the simultaneous equations \( -A + B = 2\) and \(A + B = 0\): - Adding these gives \(2B = 2 \Rightarrow B = 1\). - Substituting \(B = 1\) into \(A + B = 0\) gives \(A + 1 = 0 \Rightarrow A = -1\).6. Substitute \(A = -1\) into \(C = -A\) gives \(C = 1\).

Step 6: Write the partial fraction decomposition

Substitute \(A = -1, B = 1, C = 1, D = 0\) back into the partial fraction decomposition form:\[\frac{2x^2}{(x+1)^2(x^2+1)} = \frac{-1}{x+1} + \frac{1}{(x+1)^2} + \frac{x}{x^2+1}\]

Step 7: Integrate each term separately

Integrate each term separately:1. \(\int \frac{-1}{x+1} \, dx = -\ln|x+1| + C_1\)2. \(\int \frac{1}{(x+1)^2} \, dx = -\frac{1}{x+1} + C_2\)3. \(\int \frac{x}{x^2+1} \, dx = \frac{1}{2}\ln|x^2+1| + C_3\)

Step 8: Combine the integrals

Combine the integrals to get the final equation:\[\int \frac{2x^2}{(x+1)^2(x^2+1)} \, dx = -\ln|x+1| - \frac{1}{x+1} + \frac{1}{2} \ln|x^2+1| + C\]where \(C = C_1 + C_2 + C_3\) is the constant of integration.

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with finding the integral of functions. In simple terms, integration is the reverse process of differentiation. While differentiation splits a function into smaller parts, integration combines parts to form a whole. In calculus, there are two major types of integrals: definite and indefinite.

• Definite integrals compute the area under a curve within a specified interval.
• Indefinite integrals, like the one in our exercise, do not have specified limits, and result in a general form with a constant of integration, usually denoted as \(C\).

Understanding integration involves recognizing patterns and applying techniques like substitution, integration by parts, and decomposition to evaluate more complex expressions.

Rational Functions

Rational functions are expressions that result from the division of two polynomials. They are central in calculus due to their complex behavior and the challenges they present in integration.

• In our exercise, the rational function is represented by \( \frac{2x^2}{(x+1)^2(x^2+1)} \).
• Rational functions often require decomposition into simpler fractions, particularly when facing tricky integrations.

The method of partial fraction decomposition is frequently utilized, especially when dealing with integrals involving rational functions, to break down a complicated expression into simpler terms.

Coefficients Comparison

When performing partial fraction decomposition, finding the right coefficients is crucial. This method hinges on expressing a complicated fraction as a sum of simpler fractions. Once this is set up, coefficients comparison helps to find unknown values that satisfy the equation. It works like this:

• Expand both sides of the equation to remove the denominator.
• Collect like terms, and equate the coefficients of matching powers of \(x\) from both sides of the equation.
• Solve the resulting system of linear equations to find the values of unknown coefficients.

This method allows us to rewrite the function into its decomposed parts, facilitating simpler integration.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It's especially useful for functions that are products of algebraic and non-algebraic terms.The formula is:\[\int u \, dv = uv - \int v \, du\]where:

• \(u\) is the function chosen to be differentiated.
• \(dv\) is the function chosen to be integrated.

In the original exercise, each term of the partial fraction decomposition is integrated separately. For some terms, such as \(\frac{x}{x^2+1}\), integration by parts provides a systematic approach to evaluate the integral. Recognizing when to apply integration by parts is key to simplifying problems involving more complex integrations.