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Chapter 7: Problem 29

Evaluate the integral. \(\int \frac{2 x^{2}-1}{(4 x-1)\left(x^{2}+1\right)} d x\)

Short Answer

Expert verified
The integral is \(\frac{3}{4} \ln|4x-1| + \frac{3}{2} \ln|x^2+1| - 0.75 \tan^{-1}(x) + C\).

Step by step solution

01

Decompose the Fraction

To integrate \(\frac{2x^2-1}{(4x-1)(x^2+1)}\), we need to decompose it into partial fractions. Assume:\[\frac{2x^2-1}{(4x-1)(x^2+1)} = \frac{A}{4x-1} + \frac{Bx+C}{x^2+1}\]where \(A\), \(B\), and \(C\) are constants to be determined.
02

Clear the Denominator

Multiply both sides of the equation by \((4x-1)(x^2+1)\) to clear the denominators:\[2x^2-1 = A(x^2+1) + (Bx+C)(4x-1)\]Expand the terms on the right-hand side to get:\[2x^2-1 = Ax^2 + A + (4Bx^2 - Bx + 4Cx - C)\]Combine like terms:\[2x^2-1 = (A + 4B)x^2 + (4C - B)x + (A - C)\]
03

Match Coefficients

By equating the coefficients on both sides for corresponding powers of \(x\), we get the following system of equations:1. \(A + 4B = 2\)2. \(4C - B = 0\)3. \(A - C = -1\)
04

Solve for Constants

Solve the system of equations:From equation (2): \(B = 4C\)Substitute \(B = 4C\) into equation (1):\(A + 16C = 2\)From equation (3): \(C = A + 1\)Substitute \(C = A + 1\) into \(B = 4C\) and \(A + 16C = 2\) to find \(A\), \(B\), and \(C\).Eventually, solving gives \(A = 0.75\), \(B = 3\), \(C = -0.75\).
05

Integrate Each Term

The partial fraction decomposition gives us:\[\frac{3}{4x-1} + \frac{(3x - 0.75)}{x^2 + 1}\]Integrate each term separately:- Integral of \(\frac{3}{4x-1}\) is \(\frac{3}{4} \ln|4x-1|\).- Integral of \(\frac{3x}{x^2+1}\) is \(\frac{3}{2} \ln|x^2+1|\).- Integral of \(\frac{-0.75}{x^2+1}\) is \(-0.75 \tan^{-1}(x)\).
06

Combine Results

Combine the integrated results from each part:\[\int \frac{2x^2-1}{(4x-1)(x^2+1)} \, dx = \frac{3}{4} \ln|4x-1| + \frac{3}{2} \ln|x^2+1| - 0.75 \tan^{-1}(x) + C\]where \(C\) is the constant of integration.

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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 used to break down complex rational expressions into simpler fractions. This simplification makes the integration process more straightforward. In the given exercise, we needed to decompose \( \frac{2x^2-1}{(4x-1)(x^2+1)} \) into a sum of simpler fractions. Here's how we did it:
  • Assume that \( \frac{2x^2-1}{(4x-1)(x^2+1)} \) can be expressed as \( \frac{A}{4x-1} + \frac{Bx+C}{x^2+1} \).
  • To find constants \( A \), \( B \), and \( C \), we multiplied through by the denominator to get rid of the fractions.
  • By expanding and combining like terms, we created equations by matching coefficients for corresponding powers of \( x \). This led us to a system of equations.
  • Solving these equations gave us the values of \( A, B, \) and \( C \).
The step-by-step breakdown helps ensure that each fraction corresponds to a simpler component of the original function, making integration feasible. Mastering this technique requires practice, focus on correctly setting up initial assumptions, and efficiently solving systems of equations.
Integration Techniques
Integration techniques are methods used to find the integral, or antiderivative, of a function. In the context of our problem, we needed to integrate the simplified expressions after applying partial fraction decomposition:
  • For simple fractions like \( \frac{3}{4x-1} \), logarithmic integration is used. It results in \( \frac{3}{4} \ln|4x-1| \).
  • When integrating \( \frac{3x}{x^2+1} \), substitution is useful. Let \( u = x^2 + 1 \), which simplifies the integral and results in \( \frac{3}{2} \ln|x^2+1| \).
  • For fractions involving a constant term over a quadratic expression, like \( \frac{-0.75}{x^2+1} \), notice it matches the derivative form of the arctangent function, leading to \( -0.75 \tan^{-1}(x) \).
These techniques show that different parts of the decomposition might require distinct methods of integration. Recognizing which technique suits each fraction is crucial and can significantly ease the process.
Inverse Trigonometric Functions
Inverse trigonometric functions are useful in integration, especially when dealing with quadratic expressions in the denominator. In our problem, one of the terms \( \frac{-0.75}{x^2+1} \) fit the format of an inverse trigonometric function:
  • Recall that the integral of \( \frac{1}{x^2+1} \) is \( \tan^{-1}(x) + C \). This knowledge comes into play with minimal modifications needed.
  • The negative constant in \( \frac{-0.75}{x^2+1} \) simply scales the result to \( -0.75 \tan^{-1}(x) \).
Inverse trigonometric functions, like arctan, are instrumental because they relate algebraic expressions to angles, allowing for more complex integrals to be solved in terms of simpler geometrical concepts. Recognizing these patterns is a valuable skill in integral calculus, enabling quick solutions to determinate integrals.

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Most popular questions from this chapter

(a) Make an appropriate \(u\) -substitution of the form \(u=x^{1 / n}\) or \(u=(x+a)^{1 / n}\), and then evaluate the integral. (b) If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a). $$ \int \frac{x}{\sqrt{x+1}} d x $$

Some integrals that can be evaluated by hand cannot be evaluated by all computer algebra systems. Evaluate the integral by hand, and determine if it can be evaluated on your CAS. $$ \int \frac{1}{x^{10}+x} d x $$

Evaluate the integral. $$ \int \csc ^{4} x d x $$

Use any method to solve for \(x\). $$ \int_{2}^{x} \frac{1}{t(4-t)} d t=0.5,2

(a) Give a reasonable informal argument, based on areas, that explains why the integrals $$ \int_{0}^{+\infty} \sin x d x \text { and } \int_{0}^{+\infty} \cos x d x $$ diverge. (b) Show that \(\int_{0}^{+\infty} \frac{\cos \sqrt{x}}{\sqrt{x}} d x\) diverges.
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