/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 44

Evaluate the following integrals. $$\int \frac{x^{2}+3 x+2}{x\left(x^{2}+2 x+2\right)} d x$$

Short Answer

Expert verified
Question: Find the integral of the following function: \(\int \frac{x^2 + 3x + 2}{x(x^2 + 2x + 2)} dx\). Answer: The integral of the given function is \(\int \frac{x^2 + 3x + 2}{x(x^2 + 2x + 2)} dx = \ln |x| + \arctan(x + 1) + C\), where C is the constant of integration.

Step by step solution

01

Perform Partial Fraction Decomposition

In order to perform partial fraction decomposition, we first set the following form: $$\frac{x^2 + 3x + 2}{x(x^2 + 2x + 2)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 2x + 2}$$ Now multiply both sides by the denominator \(x(x^2 + 2x + 2)\): $$x^2 + 3x + 2 = A(x^2 + 2x + 2) + (Bx + C)x$$ The next step is to find the coefficients A, B, and C. To do so, we compare the right-hand side to the left-hand side, which gives us the following system of equations: $$A(x^2 + 2x + 2) + (Bx + C)x = x^2 + 3x + 2$$ Comparing the coefficients of \(x^2, x,\) and constants on both sides, we get: $$A + B = 1$$ $$2A + C = 3$$ $$2A = 2$$ Now, solve this system of equations for A, B, and C.
02

Solve for Coefficients A, B, and C

From the system of equations obtained in step 1, we can easily find A: $$2A = 2 \implies A = 1$$ Now, we substitute A = 1 into the other two equations and solve for B and C: $$1 + B = 1 \implies B = 0$$ $$2(1) + C = 3 \implies C = 1$$ Now we can rewrite the integrand using A, B, and C: $$\frac{x^2 + 3x + 2}{x(x^2 + 2x + 2)} = \frac{1}{x} + \frac{1}{x^2 + 2x + 2}$$
03

Integrate Each Term Separately

Now that we have simplified the integrand, we can integrate each term separately: $$\int \left(\frac{1}{x} + \frac{1}{x^2 + 2x + 2}\right) dx = \int \frac{1}{x} dx + \int \frac{1}{x^2 + 2x + 2} dx $$ The first integral can be directly integrated as: $$\int \frac{1}{x} dx = \ln |x| + C_1$$ For the second integral, we first need to complete the square in the denominator: $$x^2 + 2x + 2 = (x + 1)^2 + 1$$ Now, integrate the second term: $$\int \frac{1}{(x + 1)^2 + 1} dx$$ This integral follows the form \(\int \frac{1}{u^2 + 1} du\), which is equal to \(\arctan(u) + C_2\). Therefore, substitute \(u = x + 1\), with \(du = dx\): $$\int \frac{1}{(x + 1)^2 + 1} dx = \arctan(x + 1) + C_2$$
04

Combine the Results

Now, we combine the two results obtained in step 3: $$\int \frac{x^2 + 3x + 2}{x(x^2 + 2x + 2)} dx = \ln |x| + \arctan(x + 1) + C$$ where \(C = C_1 + C_2\) is the constant of integration. So, the final result is: $$\int \frac{x^{2}+3 x+2}{x\left(x^{2}+2 x+2\right)} d x = \ln |x| + \arctan(x + 1) + C$$

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Most popular questions from this chapter

Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral. $$\int \frac{d x}{\sqrt{x}+\sqrt[3]{x}} ; x=u^{6}$$

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Challenge Show that with the change of variables \(u=\sqrt{\tan x}\) the integral \(\int \sqrt{\tan x} d x\) can be converted to an integral amenabl to partial fractions. Evaluate \(\int_{0}^{\pi / 4} \sqrt{\tan x} d x\)
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