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Chapter 7: Problem 85

Refer to the summary box (Partial Fraction Decompositions) and evaluate the following integrals. $$\int \frac{x}{(x-1)\left(x^{2}+2 x+2\right)^{2}} d x$$

Short Answer

Expert verified
$$\int \frac{x}{x^3 + 2x^4+x^5} dx$$ Short Answer: The integral of the given function is: $$-\frac{1}{2}\ln\left| 1-\frac{1}{x} \right| +\frac{1}{4}\arctan(x+1) + \frac{1}{8} \frac{1}{x+1} + C$$

Step by step solution

01

Divide the numerator and denominator

Divide the numerator and denominator by x, so that the integrand takes the following form: $$\int \frac{1}{(\frac{x}{x}-\frac{1}{x})\left(\frac{x^2}{x}+2 \frac{x^2}{x}+2\right)^{2}} d x$$
02

Perform the partial fraction decomposition

Let's perform the partial fraction decomposition. The given integral becomes: $$\int \frac{1}{(1-\frac{1}{x})(x^2+2x+2)^2} dx$$ Now, we can write it as: $$\int \frac{A}{1-\frac{1}{x}} + \frac{Bx+C}{(x^2+2x+2)}+\frac{Dx+E}{(x^2+2x+2)^2} dx$$ where we need to find A, B, C, D, and E.
03

Solve for A, B, C, D, and E

To find the constants A, B, C, D, and E, follow these steps: 1. Clear fractions by multiplying both sides by \((x^2+2x+2)^2(1-\frac{1}{x})\). 2. Divide the resulting polynomial equation into different parts. 3. Find A, B, C, D, and E by equating the coefficients of the corresponding powers of x. After doing this, we get: $$A=-\frac{1}{2}, \qquad B=0, \qquad C=\frac{1}{4}, \qquad D=0, \qquad E=-\frac{1}{4}$$
04

Rewrite the integral with constants

Now that we have the constants, let's rewrite the given integral with these constants: $$\int -\frac{1}{2}\frac{dx}{1-\frac{1}{x}} +\frac{1}{4}\frac{dx}{(x^2+2x+2)} -\frac{1}{4}\frac{dx}{(x^2+2x+2)^2}$$
05

Integrate the simpler fractions

Let's integrate each simpler fraction individually: 1. \(\int -\frac{1}{2}\frac{dx}{1-\frac{1}{x}}\): Substitute \(u=1-\frac{1}{x}\), then integrate. 2. \(\int \frac{1}{4}\frac{dx}{(x^2+2x+2)}\): Complete the square in the denominator, then use the substitution \(u=x+1\). 3. \(\int -\frac{1}{4}\frac{dx}{(x^2+2x+2)^2}\): Complete the square in the denominator and use the substitution \(t=x+1\). After integrating the fractions, we get: $$-\frac{1}{2}\ln\left| 1-\frac{1}{x} \right| +\frac{1}{4}\arctan(x+1) + \frac{1}{8} \frac{1}{x+1} + C$$ Now, combining all the fractions, we get our final answer: $$-\frac{1}{2}\ln\left| 1-\frac{1}{x} \right| +\frac{1}{4}\arctan(x+1) + \frac{1}{8} \frac{1}{x+1} + C$$

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Most popular questions from this chapter

On the interval \([0,2],\) the graphs of \(f(x)=x^{2} / 3\) and \(g(x)=x^{2}\left(9-x^{2}\right)^{-1 / 2}\) have similar shapes. a. Find the area of the region bounded by the graph of \(f\) and the \(x\) -axis on the interval [0,2] b. Find the area of the region bounded by the graph of \(g\) and the \(x\) -axis on the interval [0,2] c. Which region has the greater area?

When is the volume finite? Let \(R\) be the region bounded by the graph of \(f(x)=x^{-p}\) and the \(x\) -axis, for \(x \geq 1.\) a. Let \(S\) be the solid generated when \(R\) is revolved about the \(x\) -axis. For what values of \(p\) is the volume of \(S\) finite? b. Let \(S\) be the solid generated when \(R\) is revolved about the \(y\) -axis. For what values of \(p\) is the volume of \(S\) finite?

Suppose that a function \(f\) has derivatives of all orders near \(x=0 .\) By the Fundamental Theorem of Calculus, \(f(x)-f(0)=\int_{0}^{x} f^{\prime}(t) d t\) a. Evaluate the integral using integration by parts to show that $$f(x)=f(0)+x f^{\prime}(0)+\int_{0}^{x} f^{\prime \prime}(t)(x-t) d t.$$ b. Show (by observing a pattern or using induction) that integrating by parts \(n\) times gives $$\begin{aligned} f(x)=& f(0)+x f^{\prime}(0)+\frac{1}{2 !} x^{2} f^{\prime \prime}(0)+\cdots+\frac{1}{n !} x^{n} f^{(n)}(0) \\ &+\frac{1}{n !} \int_{0}^{x} f^{(n+1)}(t)(x-t)^{n} d t+\cdots \end{aligned}$$ This expression is called the Taylor series for \(f\) at \(x=0\).

Evaluate the following integrals. Consider completing the square. $$\int_{2+\sqrt{2}}^{4} \frac{d x}{\sqrt{(x-1)(x-3)}}$$

\(A\) 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 that s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated: $$ F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1} $$ Verify the following Laplace transforms, where a is a real number. $$f(t)=t \longrightarrow F(s)=\frac{1}{s^{2}}$$
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