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Integrate by hand and check your answers using a CAS. $$\int \frac{d x}{16 x^{3}-4 x^{2}+4 x-1}$$

Short Answer

Expert verified
The integral is \( \frac{1}{12} \ln |4x - 1| - \frac{1}{12} \arctan(2x) + C \).

Step by step solution

01

Identify the integral type

The given integral is \( \int \frac{d x}{16 x^{3}-4 x^{2}+4 x-1} \). This is a rational function, which suggests using partial fraction decomposition to solve it.
02

Factor the denominator

Check if the denominator \( 16x^3 - 4x^2 + 4x - 1 \) can be factored. Using the Rational Root Theorem and synthetic division, we find the factorization: \((4x - 1)(4x^2 + 1)\). Then the integral becomes: \( \int \frac{d x}{(4x - 1)(4x^2 + 1)} \).
03

Set up partial fractions

Express the integrand as partial fractions: \( \frac{1}{(4x - 1)(4x^2 + 1)} = \frac{A}{4x - 1} + \frac{Bx + C}{4x^2 + 1} \).
04

Solve for coefficients

Multiply through by the common denominator \((4x - 1)(4x^2 + 1)\) and equate coefficients to solve for \(A\), \(B\), and \(C\). By performing polynomial expansion and matching coefficients, solve for \(A = 1/3\), \(B = 0\), \(C = -1/3\).
05

Integrate each term

Integrate each of the partial fractions: \(\int \frac{1/3}{4x - 1} dx\) and \(\int \frac{-1/3}{4x^2 + 1} dx\). The first integral evaluates to \( \frac{1}{12} \ln |4x - 1| \). The second integral evaluates to \(-\frac{1}{3} \times \frac{1}{4} \times \arctan(2x)\) or \(-\frac{1}{12} \arctan(2x)\).
06

Combine results and finalize solution

Combine the results of the integrals: \( \int \frac{d x}{16 x^{3}-4 x^{2}+4 x-1} = \frac{1}{12} \ln |4x - 1| - \frac{1}{12} \arctan(2x) + C \).
07

Verify with a CAS

Check the solution using a Computer Algebra System (CAS) to ensure it matches the derived solution. Upon comparison, the CAS confirms the computation: \( \frac{1}{12} \ln |4x - 1| - \frac{1}{12} \arctan(2x) + C \).

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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 to express a complex fraction into simpler fractions that are easier to integrate. It is particularly useful for rational functions, which are ratios of two polynomials.
This method involves breaking down the original function using algebraic fractions, prior to integration.Let's consider the example at hand: - We begin with the integrand \( \frac{1}{(4x - 1)(4x^2 + 1)} \).
Here, the goal is to represent it as a sum of simpler fractions: \( \frac{A}{4x - 1} + \frac{Bx + C}{4x^2 + 1} \).This new representation helps in performing the integration term by term. Each term in the decomposition is easier to integrate, reducing the complexity of the original integral.
The coefficients \(A\), \(B\), and \(C\) can be determined by clearing the denominators and equating coefficients of powers of \(x\).
For this particular case, once you solve for the coefficients, you have \(A = \frac{1}{3}\), \(B = 0\), and \(C = -\frac{1}{3}\).Partial fraction decomposition simplifies the integration process considerably, especially when working with higher-degree polynomials.
Integration Techniques
Once we have decomposed the fraction, it's time to integrate each component separately.For this exercise, we perform the following:- Integrate \(\frac{1/3}{4x - 1}\) with respect to \(x\). This is a straightforward integration that results in \( \frac{1}{12} \ln |4x - 1|\).- Next, integrate \(\frac{-1/3}{4x^2 + 1}\) with respect to \(x\). This requires knowledge of the integral of \(\tan^{-1}(x)\), leading to \( -\frac{1}{12} \arctan(2x)\).These techniques often involve using basic integration formulas. Remember, the natural logarithm and inverse trigonometric functions frequently result from rational function integration. Putting the results together, the integral of the original function becomes:- \( \frac{1}{12} \ln |4x - 1| - \frac{1}{12} \arctan(2x) + C \), where \(C\) is the constant of integration.
By separating the integrand into smaller parts, integration becomes a matter of applying familiar, fundamental antiderivatives.
Polynomial Factorization
Polynomial factorization is the process of breaking down a polynomial equation into simpler polyomials which can multiply to give the original polynomial.
This is a crucial step in applying partial fraction decomposition, as it allows us to identify distinct, simpler terms.In our exercise, we needed to factor \(16x^3 - 4x^2 + 4x - 1\).
Factorization can sometimes be tricky, but tools like the Rational Root Theorem and synthetic division are quite helpful.
By applying these techniques, the original polynomial was successfully factored into \((4x - 1)(4x^2 + 1)\).With the polynomial expressed in factored form, it becomes possible to set up our partial fraction decomposition.
Proper factorization is essential, as an incorrect factorization would lead to incorrect partial fractions, which could throw off our entire integration process.
Remember to verify your factorization with a couple of test values or a graphing tool if available as a cross-check.

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Most popular questions from this chapter

The region bounded below by the \(x\) -axis and above by the portion of \(y=\sin x\) from \(x=0\) to \(x=\pi\) is revolved about the \(x\) -axis. Find the volume of the resulting solid.

Find the arc length of the curve \(y=\ln (\cos x)\) over the interval \([0, \pi / 4]\)

What is "improper" about an integral over an infinite interval? Explain why Definition 5.5 .1 for \(\int_{a}^{b} f(x) d x\) fails for \(\int_{a}^{+\infty} f(x) d x .\) Discuss a strategy for assigning a value to \(\int_{a}^{+\infty} f(x) d x\).

Suppose that during a period \(t_{0} \leq t \leq t_{1}\) years, a company has a continuous income stream at a rate of \(I(t)\) dollars per year at time \(t\) and that this income is invested at an annual rate of \(r \%,\) compounded continuously. The value (in dollars) of this income stream at the end of the time period \(t_{0} \leq t \leq t_{1},\) called the stream's future value, can be calculated using $$F V=\int_{t_{0}}^{t_{1}} I(t) e^{r\left(t_{1}-t\right)} d t$$ The present value (in dollars) of the income stream is given by $$P V=\int_{t_{0}}^{t_{1}} I(t) e^{-r\left(t-t_{0}\right)} d t$$ The present value is the amount that, if put in the bank at time \(t=t_{0}\) at \(r \%\) compounded continuously, with no additional deposits, would result in a balance of \(F V\) dollars at time \(t=t_{1}\) That is, $$F V=P V \times e^{r\left(t_{1}-t_{0}\right)}$$ In each exercise, (a) find the future value \(F V\) for the given income stream \(I(t)\) and interest rate \(r\) and time period \(t_{0} \leq t \leq t_{1}\) (b) find the present value \(P V\) of the income stream over the time period; and (c) verify that \(F V\) and \(P V\) satisfy the relationship given above. $$I(t)=2000 t+400 e^{-t} ; r=8 \% ; 0 \leq t \leq 10$$

Later in the text, we will show that $$\int_{0}^{+\infty} e^{-x^{2}} d x=\frac{1}{2} \sqrt{\pi}$$ Confirm that this is reasonable by using a CAS or a calculator with a numerical integration capability.

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