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$$\text {Evaluate the following integrals.}$$ $$\int \frac{2 x^{2}+5 x+5}{(x+1)\left(x^{2}+2 x+2\right)} d x$$

Short Answer

Expert verified
Question: Find the integral of the given function: $$\int \frac{2 x^{2}+5 x+5}{(x+1)\left(x^{2}+2 x+2\right)} d x$$ Answer: $$\int \frac{2 x^{2}+5 x+5}{(x+1)\left(x^{2}+2 x+2\right)} d x = \ln|x+1|+\ln|x^2+2x+2|+C$$

Step by step solution

01

Perform partial fraction decomposition

In this step, we decompose the given fraction into simpler fractions which can then be easily integrated. We have the following integral: $$\int \frac{2 x^{2}+5 x+5}{(x+1)\left(x^{2}+2 x+2\right)} d x$$ Let's perform partial fraction decomposition and express the integrand as: $$\frac{2x^2 +5x + 5}{(x+1)(x^2 +2x+2)} = \frac{A}{x+1}+\frac{Bx+C}{x^2+2x+2}$$ Now, we want to find the values of A, B, and C.
02

Clear the denominator and simplify

To obtain the values of A, B, and C, we clear the denominator by multiplying both sides by \((x+1)(x^2 +2x+2)\), and we get: $$(2x^2 +5x + 5) = A(x^2+2x+2)+ (Bx+C)(x+1)$$
03

Solve for A, B, and C

Now we should find the expressions for A, B, and C by comparing coefficients on both sides. Let's calculate these values: By comparing the coefficients of \(x^2\) terms, we get: \(2 = A+B\) By comparing the coefficients of \(x\) terms, we get: \(5 = 2A + B + C\) By comparing constant terms, we get: \(5 = 2A+C\) Solving this system of linear equations, we find out that A = 1, B = 1, and C = 1. Thus, we can rewrite our integrand as: $$\frac{1}{x+1} + \frac{x+1}{x^2 + 2x + 2}$$ Now, we can integrate these smaller fractions separately:
04

Integration

Integrating each term separately: $$\int \frac{1}{x+1}dx + \int \frac{x+1}{x^2 + 2x + 2}dx$$ The first integral can be directly integrated using basic integration rules: $$\int \frac{1}{x+1}dx = \ln|x+1|+ C_1$$ For the second integral, we use substitution: Let \(u = x^2 + 2x + 2\). Then, \(du = (2x + 2)dx\) So, the second integral becomes: $$\int \frac{1}{u} du = \ln|u|+C_2 = \ln|x^2+2x+2|+C_2$$ Now, let's combine both parts to get the final answer.
05

Combine the results

Adding the results of both integrals and the constants, we have: $$\ln|x+1|+\ln|x^2+2x+2|+C$$ Where \(C = C_1 + C_2\) is the constant of integration. The final answer is: $$\int \frac{2 x^{2}+5 x+5}{(x+1)\left(x^{2}+2 x+2\right)} d x = \ln|x+1|+\ln|x^2+2x+2|+C$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration Techniques
Mastering integration techniques is crucial for solving complex calculus problems. One such technique is partial fraction decomposition, which simplifies the integration process by breaking a complex rational expression into simpler fractions that can be integrated individually. This is particularly useful when dealing with integrals of rational functions.

For example, the integral \(\int \frac{2 x^{2}+5 x+5}{(x+1)(x^{2}+2 x+2)} d x\) can appear daunting at first. However, by applying partial fraction decomposition, we can express the integrand as the sum of simpler fractions, each with its own unique coefficient, which can then be integrated using basic integration rules or appropriate substitutions. This technique often involves solving a system of linear equations to find the coefficients, as seen in the initial stages of the provided exercise.
Integral Calculus
Integral calculus is a branch of mathematics focused on the accumulation of quantities and the areas under and between curves. When we integrate a function, we're essentially finding the total sum of infinitesimally small data points. This process often requires familiarity with a wide range of functions and their antiderivatives.

For instance, in the solution provided, we start by computing the antiderivative of simple functions like \(\frac{1}{x+1}\) and \(\frac{x+1}{x^2 + 2x + 2}\). The latter requires a substitution method to simplify the integrand to a more familiar form. In this case, substituting the quadratic expression with \(u\) simplifies the integral to a natural logarithmic function. This illustrates how integral calculus often involves strategic manipulation of an equation to reduce it to a basic form that can be integrated.
Linear Equations System
A linear equations system is a collection of two or more linear equations involving the same set of variables. Solving these systems is a fundamental skill in algebra and is especially useful in partial fraction decomposition, where we need to determine the unknown coefficients of the decomposed fractions.

In our exercise example, we created a system of equations by equating coefficients from both sides of an identity and then solved for the unknowns A, B, and C. This procedure demonstrates how crucial it is to develop a strong understanding of how to maneuver through linear systems, as it's often the gateway to advancing in more complex areas of mathematics, like calculus.

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Most popular questions from this chapter

Trapezoid Rule and concavity Suppose \(f\) is positive and its first two derivatives are continuous on \([a, b] .\) If \(f^{\prime \prime}\) is positive on \([a, b]\) then is a Trapezoid Rule estimate of \(\int_{a}^{b} f(x) d x\) an underestimate or overestimate of the integral? Justify your answer using Theorem 8.1 and an illustration.

A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$$ where we assume s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated using integration by parts: $$F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1}$$ Verify the following Laplace transforms, where a is a real number. $$f(t)=1 \rightarrow F(s)=\frac{1}{s}$$

Show that \(\int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} d x=\pi\) in the following steps. a. Integrate by parts with \(u=\sqrt{x} \ln x\). b. Change variables by letting \(y=1 / x\). c. Show that \(\int_{0}^{1} \frac{\ln x}{\sqrt{x}(1+x)} d x=-\int_{1}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x\) (and that both integrals converge). Conclude that \(\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x=0\). d. Evaluate the remaining integral using the change of variables \(z=\sqrt{x}\). (Source: Mathematics Magazine \(59,1,\) Feb 1986 )

sine integral The theory of diffraction produces the sine integral function \(\mathrm{Si}(x)=\int_{0}^{x} \frac{\sin t}{t} d t .\) Use the Midpoint Rule to approximate \(\left.\operatorname{Si}(1) \text { and } \operatorname{Si}(10) . \text { (Recall that } \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 .\right)\) Experiment with the number of subintervals until you obtain approximations that have an error less than \(10^{-3}\). A rule of thumb is that if two successive approximations differ by less than \(10^{-3}\), then the error is usually less than \(10^{-3} .\)

Let \(a>0\) and \(b\) be real numbers. Use integration to confirm the following identities. (See Exercise 73 of Section 8.2) a. \(\int_{0}^{\infty} e^{-a x} \cos b x d x=\frac{a}{a^{2}+b^{2}}\) b. \(\int_{0}^{\infty} e^{-a x} \sin b x d x=\frac{b}{a^{2}+b^{2}}\)

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