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Magazine Chapter 11: Problem 18
Find the indefinite integral. $$\int \frac{x^{2}-1}{x^{3}-3 x+1} d x$$
Short Answer
Expert verified
\(\int \frac{x^2-1}{x^3-3x+1} dx =\frac{\sqrt{5}-1}{2\sqrt{5}} \ln |x^2 - (\frac{3 - \sqrt{5}}{2})| + C\)
Step by step solution
01
1. Setup
Identify the given integral function:
\(\int \frac{x^2 - 1}{x^3 - 3x + 1} dx \)
02
2. Simplify the integrand
Observe the numerator \(x^2 - 1\), which can be factored to \((x-1)(x+1)\).
So, the integral becomes:
\(\int \frac{(x-1)(x+1)}{x^3 - 3x + 1} dx\)
03
3. Consider a substitution
Since we have quadratic function in the numerator and cubic in the denominator, we may try a substitution technique. Take \(\boldsymbol{u = x^2}\) and then find the differential of \(u\) with respect to \(x\):
\(\boldsymbol{du = 2x dx}\)
Now try to arrange the expression in terms of \(u\), and solve the integral with the substitution:
\( \int \frac{u - 1}{u^2 - 3u + 1} \frac{du}{2} \)
04
4. Solve the substituted expression
Now, let's integrate the new expression:
\( \frac{1}{2} \int \frac{u - 1}{u^2 - 3u + 1} du \)
To move forward with this integral, try to perform partial fractions decomposition.
05
5. Perform partial fractions decomposition
The decomposition of the fraction can be written as:
\(\frac{u - 1}{u^2 - 3u + 1} = \frac{A}{u - a} + \frac{B}{u - b}\)
where \(a\) and \(b\) are the roots of the denominator, and \(A\) and \(B\) are constants to be determined. Notice that for this problem, the denominator can be factored as:
\((u - a)(u - b)\)
Find the roots for this quadratic equation by solving \(u^2 - 3u + 1 = 0\), which are \(a = \frac{3 + \sqrt{5}}{2}\) and \(b = \frac{3 - \sqrt{5}}{2}\). So,
\(\frac{u - 1}{(u - (\frac{3 + \sqrt{5}}{2}))(u - (\frac{3 - \sqrt{5}}{2}))} = \frac{A}{u - (\frac{3 + \sqrt{5}}{2})} + \frac{B}{u - (\frac{3 - \sqrt{5}}{2})}\)
06
6. Calculate A and B
To find the constants \(A\) and \(B\), first clear the fractions:
\((u-1) = A(u - (\frac{3 - \sqrt{5}}{2})) + B(u - (\frac{3 + \sqrt{5}}{2}))\)
Next, plug in the roots for \(u\) to solve for one constant at a time:
For \(u = (\frac{3 + \sqrt{5}}{2})\) we have:
\((\frac{\sqrt{5}-1}{2}) = B((\frac{2\sqrt{5}}{2}))\)
\(A = 0\) and \(B = \frac{\sqrt{5}-1}{\sqrt{5}}\)
07
7. Rewrite the integrand and integrate
With the partial fractions decomposition done, rewrite the integrand and integrate:
\(\frac{1}{2} \int \frac{u - 1}{u^2 - 3u + 1} du = \frac{1}{2} \int \frac{\frac{\sqrt{5}-1}{\sqrt{5}}}{u - (\frac{3 - \sqrt{5}}{2})} du\)
Now integrate the function:
\(\frac{1}{2} \int \frac{\frac{\sqrt{5}-1}{\sqrt{5}}}{u - (\frac{3 - \sqrt{5}}{2})} du = \frac{\sqrt{5}-1}{2\sqrt{5}} \ln |u - (\frac{3 - \sqrt{5}}{2})| + C\)
08
8. Substitute back x
Now, remember that \(u = x^2\). Substitute back \(x\):
\(\frac{\sqrt{5}-1}{2\sqrt{5}} \ln |x^2 - (\frac{3 - \sqrt{5}}{2})| + C\)
So, the indefinite integral of the given function is:
\(\int \frac{x^2-1}{x^3-3x+1} dx =\frac{\sqrt{5}-1}{2\sqrt{5}} \ln |x^2 - (\frac{3 - \sqrt{5}}{2})| + C\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Substitution
Integration by substitution is a technique used to find integrals by transforming them into simpler forms. This method is particularly useful when dealing with composite functions. The basic idea is to replace a portion of the integral with a single variable, simplifying the integrand.
For the given problem, we substituted \( u = x^2 \) because the function we are integrating involves \( x^2 \), which simplifies our function ∶ \
For the given problem, we substituted \( u = x^2 \) because the function we are integrating involves \( x^2 \), which simplifies our function ∶ \
- Transform the integral into a function of \( u \).
- Compute the differential, \( du = 2x \, dx \), to replace \( dx \) in the integrand.
- Solve the integral in terms of \( u \), and later replace \( u \) back to \( x^2 \).
Partial Fraction Decomposition
Partial fraction decomposition is a powerful tool for integrating rational functions. It involves breaking down a complicated fraction into simpler parts, usually as a sum of fractions. Each fraction will have a simpler denominator, making it easier to integrate individually.
In our problem, we used partial fraction decomposition on the expression \( \frac{u-1}{u^2-3u+1} \) to simplify it.
In our problem, we used partial fraction decomposition on the expression \( \frac{u-1}{u^2-3u+1} \) to simplify it.
- First, the denominator \( u^2 - 3u + 1 \) was factored into linear components.
- Next, we expressed the complicated fraction as a sum of two simple fractions: \( \frac{A}{u - a} + \frac{B}{u - b} \).
- We determined the constants \( A \) and \( B \) by solving equations derived from the roots of the denominator.
Factoring Polynomials
Factoring polynomials is a central algebraic skill used in integration. By breaking down a polynomial expression into its factors, we can often simplify the integral better. Factoring helps especially when dealing with complex rational expressions.
In the exercise, we factored both the numerator \( x^2 - 1 \) into \( (x-1)(x+1) \) and the variable \( u \) expression in the substitution step \( u^2 - 3u + 1 \) into its respective linear terms.
In the exercise, we factored both the numerator \( x^2 - 1 \) into \( (x-1)(x+1) \) and the variable \( u \) expression in the substitution step \( u^2 - 3u + 1 \) into its respective linear terms.
- The numerator \( x^2 - 1 \) being a difference of squares was smoothly factored.
- The quadratic \( u^2 - 3u + 1 \) was factored using the quadratic formula to find its roots, facilitating the decomposition in the next stage.
