/*! 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
Based on the step-by-step solution, provide a short answer to the problem: Given the integral $$\int \frac{x+2}{x^2+4} d x$$, we first make the substitution \(u = x^2 + 4\) and find the differential \(d u\). Then, we substitute the expressions for x and dx in the integral. After simplifying and dividing the integral into parts, we compute the integral and get $$\frac{1}{4}\ln |x^2+4| + \ln |x^2+4| + C\ = \frac{5}{4}\ln |x^2+4| + C\). Finally, the evaluated integral is $$\int \frac{x+2}{x^2+4} d x = \frac{5}{4}\ln |x^2+4| + C$$.

Step by step solution

01

Let u=x^2+4 and find du

Let's make the substitution \(u=x^2+4\). Taking the derivative of both sides with respect to x, we have: $$\frac{d u}{d x}=2 x$$ Which means, $$d u=2 x d x$$
02

Substitute x and dx in the integral

Next, we want to substitute the u-expression for the x-expression in the integral, as well as substitute du for dx: $$\int \frac{x+2}{x^2+4} d x=\int \frac{1}{2}\frac{(x+2)}{(x^2+4)}(2x d x)$$ Now, substitute \(u\) and \(d u\): $$\int \frac{1}{2}\frac{(x+2)}{u}du$$
03

Simplify and divide in parts

Now, split the integral into two separate integrals: $$\frac{1}{2}\int \frac{x}{u} d u + \frac{1}{2}\int \frac{2}{u} d u$$ Remember that \(d u = 2x d x\), so we can substitute back for \(x\): $$\frac{1}{2}\int \frac{\frac{1}{2} d u}{u} + \frac{1}{2}\int \frac{2}{u} d u$$ Now, the integral looks simpler, and we can solve each part: $$\frac{1}{4}\int \frac{d u}{u} + \int \frac{d u}{u}$$
04

Integrate

The above integrals are both of the natural logarithm form; therefore, we can write them as: $$\frac{1}{4}\ln |u| + \ln |u| +C$$ Where \(C\) is the constant of integration.
05

Substitute back for u and simplify

Now, substitute back the expression for \(u\): $$\frac{1}{4}\ln |x^2+4| + \ln |x^2+4|+C$$ Now, notice that we can combine the two logarithms using the logarithm property \(\ln a+\ln b=\ln(ab)\): $$\left(\frac{1}{4}+1\right)\ln |x^2+4|+C=\frac{5}{4}\ln |x^2+4|+C$$ Thus, the evaluated integral is: $$\int \frac{x+2}{x^2+4} d x =\frac{5}{4}\ln |x^2+4|+C$$

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Most popular questions from this chapter

\(A\) 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. $$f(t)=\cos a t \longrightarrow F(s)=\frac{s}{s^{2}+a^{2}}$$

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Many methods needed 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 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, No. 1 (February 1986): 49).

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