/*! 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 7 Find the indefinite integral. ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 7

Find the indefinite integral. $$ \int \frac{x^{2}+2 x+3}{x^{3}+3 x^{2}+9 x} d x $$

Short Answer

Expert verified
The indefinite integral of \[ \frac{x^{2}+2x+3}{x^{3}+3x^{2}+9x} dx \] is \[ log|x| + \frac{8}{3} arctan(\frac{2}{3}(x+\frac{3}{2})) + C \]

Step by step solution

01

Analyze and simplify the integral if possible

First, look at the given integral: \[ \int \frac{x^{2}+2x+3}{x^{3}+3x^{2}+9x} dx \]. It's a rational function, where the numerator is a quadratic polynomial and the denominator is a cubic polynomial. The degree of the numerator is less than the degree of the denominator, therefore we cannot do any simplification here.
02

Use partial fractions decomposition

Because the degree of the denominator is more than the numerator, we can use the method of partial fractions to simplify this integral. But first factorize the denominator, it is already in its simplest form. As per the partial fractions method, we can rewrite our integral as \[ \int \frac{x^{2}+2x+3}{x(x^{2}+3x+9)} dx \] = \[ \int \frac{A}{x} + \frac{Bx+C}{x^{2}+3x+9} dx \]. Now, we need to find the constants A, B, and C.
03

Find the constants

To find the constants A, B and C, we equate \[ \frac{x^{2}+2x+3}{x(x^{2}+3x+9)} \] = \[ \frac{A}{x} + \frac{Bx+C}{x^{2}+3x+9} \]. Then get rid of the fractions and equate coefficients on both sides: A*(x^2+3x+9) + x*(Bx+C) = x^2 + 2x + 3. This yields A = 1, B = 0 and C = 2. Thus, the integral expression becomes: \[ \int \frac{1}{x} dx + \int \frac{2}{x^{2}+3x+9} dx \].
04

Integrate

Now that we have broken down our original integral into two simpler integrals, let's solve these individually. The first is a simple log function. The integral of \[ \int \frac{1}{x} dx \] is \[ log|x| + C1 \]. The second part requires completing the square in the denominator to allow it to be expressed in form of standard integral formulas. Completing the square \[ x^{2}+3x+9 \] becomes \[ (x+\frac{3}{2})^{2} + \frac{18-9}{4} \] = \[ (x+\frac{3}{2})^{2}+\frac{9}{4} \]. Therefore, \[ \int \frac{2}{x^{2}+3x+9} dx \] becomes \[ \int \frac{2}{(x+\frac{3}{2})^{2}+\frac{9}{4}} dx \]. This is now in a more recognizable form that can be expressed as arctan function.
05

Apply Arctan substitution

To solve this integral, we use the arctan substitution method. The specific integral formula used for this substitution is: \[ \int \frac{1}{a^{2}+x^{2}} dx = \frac{1}{a} arctan(\frac{x}{a}) + C \] which simplifies to: \[ \frac{8}{3} arctan(\frac{2}{3}(x+\frac{3}{2})) + C2 \].
06

Combine the results

Now add both integration results together, combine those arbitrary constants C1 and C2 into single constant C.

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Most popular questions from this chapter

(a) integrate to find \(F\) as a function of \(x\) and (b) demonstrate the Second Fundamental Theorem of Calculus by differentiating the result in part (a). $$ F(x)=\int_{\pi / 3}^{x} \sec t \tan t d t $$

Verify the differentiation formula. \(\frac{d}{d x}[\operatorname{sech} x]=-\operatorname{sech} x \tanh x\)

Show that \(\arctan (\sinh x)=\arcsin (\tanh x)\).

Use the Second Fundamental Theorem of Calculus to find \(F^{\prime}(x)\). $$ F(x)=\int_{1}^{x} \sqrt[4]{t} d t $$

Linear and Quadratic Approximations In Exercises 33 and 34 use a computer algebra system to find the linear approximation \(P_{1}(x)=f(a)+f^{\prime}(a)(x-a)\) and the quadratic approximation \(P_{2}(x)=f(a)+f^{\prime}(a)(x-a)+\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}\) of the function \(f\) at \(x=a\). Use a graphing utility to graph the function and its linear and quadratic approximations. \(f(x)=\tanh x, \quad a=0\)
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