/*! 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 23 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 23

Evaluate the following integrals. $$\int \frac{x+2}{x^{2}+4} d x$$

Short Answer

Expert verified
Question: Evaluate the integral $\int \frac{x+2}{x^2+4} dx$. Answer: $\int \frac{x+2}{x^2+4} dx = \frac{1}{4} \left(x + \frac{C_2}{x}\right) + \ln{|x^2 + 4|} + C$.

Step by step solution

01

Choose an appropriate substitution

Let us make the substitution \(u = x^2 + 4\). Then, differentiate both sides with respect to \(x\) to obtain the differential: $$du = 2x dx$$.
02

Rewrite the integral using substitution

Now, we can substitute the \(u\) and \(du\) back into the integral: $$\int \frac{x+2}{x^2 + 4} dx = \int \frac{1}{2} \frac{(x+2)}{u}du$$.
03

Integrate with respect to u

Now, the integral is simpler and can be separated into the sum of two integrals: $$\int \frac{1}{2} \frac{(x+2)}{u}du = \frac{1}{2} \int \frac{x}{u} du + \frac{1}{2} \int \frac{2}{u} du$$. Since \(u=x^2+4\) and \(du=2x dx\), we can substitute back for \(x\): $$\frac{1}{2} \int \frac{x}{u} du = \frac{1}{2} \int \frac{1}{2x} du$$. Now we can factor out constants and integrate: $$\frac{1}{2} \int \frac{1}{2x} du + \frac{1}{2} \int \frac{2}{u} du = \frac{1}{4x} \int du + \ln{|u|} + C_1$$.
04

Substitute back for the original variable

Now, let's substitute back for the original variable \(x\): $$\frac{1}{4x} \int du + \ln{|u|} + C_1 = \frac{1}{4x} \int du + \ln{|x^2 + 4|} + C_1$$. Finally, we should integrate the term with \(\int du\): $$\frac{1}{4x} \int du + \ln{|x^2+ 4|} + C_1 = \frac{1}{4x} (u+C_2) + \ln{|x^2 + 4|} + C_1$$. Since \(u = x^2 + 4\), we can substitute again: $$\frac{1}{4x} (x^2+4+C_2) + \ln{|x^2 + 4|} + C_1 = \frac{1}{4} (x + \frac{C_2}{x}) + \ln{|x^2 + 4|} + C_1$$.
05

Simplify the result

Now, we can combine constants \(C_1\) and \(C_2\) in the final answer: $$\int \frac{x+2}{x^2+4} dx = \frac{1}{4} \left(x + \frac{C_2}{x}\right) + \ln{|x^2 + 4|} + C$$. Thus, the evaluated integral is $$\int \frac{x+2}{x^2+4} dx = \frac{1}{4} \left(x + \frac{C_2}{x}\right) + \ln{|x^2 + 4|} + C$$.

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Most popular questions from this chapter

Powers of sine and cosine It can be shown that \(\int_{0}^{\pi / 2} \sin ^{n} x d x=\int_{0}^{\pi / 2} \cos ^{n} x d x=\) \(\left\\{\begin{array}{ll}\frac{1 \cdot 3 \cdot 5 \cdots(n-1)}{2 \cdot 4 \cdot 6 \cdots n} \cdot \frac{\pi}{2} & \text { if } n \geq 2 \text { is an even integer } \\ \frac{2 \cdot 4 \cdot 6 \cdots(n-1)}{3 \cdot 5 \cdot 7 \cdots n} & \text { if } n \geq 3 \text { is an odd integer. }\end{array}\right.\)

A remarkable integral It is a fact that \(\int_{0}^{\pi / 2} \frac{d x}{1+\tan ^{m} x}=\frac{\pi}{4}\) for all real numbers \(m .\) a. Graph the integrand for \(m=-2,-3 / 2,-1,-1 / 2,0,1 / 2\) \(1,3 / 2,\) and \(2,\) and explain geometrically how the area under the curve on the interval \([0, \pi / 2]\) remains constant as \(m\) varies. b. Use a computer algebra system to confirm that the integral is constant for all \(m.\)

Evaluate the following integrals or state that they diverge. $$\int_{-2}^{6} \frac{d x}{\sqrt{|x-2|}}$$

Consider the general first-order initial value problem \(y^{\prime}(t)=a y+b, y(0)=y_{0},\) for \(t \geq 0,\) where \(a, b,\) and \(y_{0}\) are real numbers. a. Explain why \(y=-b / a\) is an equilibrium solution and corresponds to horizontal line segments in the direction field. b. Draw a representative direction field in the case that \(a>0\). Show that if \(y_{0}>-b / a,\) then the solution increases for \(t \geq 0\) and if \(y_{0}<-b / a,\) then the solution decreases for \(t \geq 0\). c. Draw a representative direction field in the case that \(a<0\). Show that if \(y_{0}>-b / a,\) then the solution decreases for \(t \geq 0\) and if \(y_{0}<-b / a,\) then the solution increases for \(t \geq 0\).

Given a Midpoint Rule approximation \(M(n)\) and a Trapezoid Rule approximation \(T(n)\) for a continuous function on \([a, b]\) with \(n\) subintervals, show that \(T(2 n)=(T(n)+M(n)) / 2\).
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