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Chapter 7: Problem 15

Evaluate the following integrals. $$\int \frac{e^{x}}{e^{x}-2 e^{-x}} d x$$

Short Answer

Expert verified
Answer: The evaluated integral of the function \(\int \frac{e^x}{e^x - 2e^{-x}} dx\) is \(\frac{1}{2} \ln |e^{2x} - 2| + C\).

Step by step solution

01

Substitution

Let's substitute \(u=e^x\). Then, we have \(du = e^x dx\). This substitution allows us to rewrite the integral in terms of \(u\): $$\int \frac {e^x}{e^x - 2e^{-x}} dx = \int \frac{u}{u - 2/u} \frac{1}{u} du$$
02

Simplify Integral

Now we have \(\int \frac{u}{u^2-2} du\). This is a rational function with a linear term in the numerator and a quadratic term in the denominator.
03

Long Division or Partial Fractions

Typically, we would use long division or partial fractions, but this integral is already in a suitable form to be integrated. Thus, we proceed to the next step.
04

Integrate

We'll integrate the rational function by using the substitution method once again: Let \(v = u^2 -2\), then \(dv = 2u du\). We then have \(\frac{1}{2} dv = u du\), and can rewrite our integral: $$\int \frac{1}{2} \frac{1}{v} dv$$ Now integrate: $$\frac{1}{2} \int \frac{1}{v} dv = \frac{1}{2} \ln |v| + C$$
05

Back-substitute for \(v\)

Replace \(v\) with the original expression: $$\frac{1}{2} \ln |u^2 - 2| + C$$
06

Back-substitute for \(u\)

Replace \(u\) with the original expression, \(u = e^x\): $$\frac{1}{2} \ln |e^{2x} - 2| + C$$ So, the evaluated integral is: $$\int \frac{e^{x}}{e^{x}-2 e^{-x}} d x = \frac{1}{2} \ln |e^{2x} - 2| + C$$

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Most popular questions from this chapter

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1-\cos x}$$

Recall that the substitution \(x=a \sec \theta\) implies that \(x \geq a\) (in which case \(0 \leq \theta<\pi / 2\) and \(\tan \theta \geq 0\) ) or \(x \leq-a\) (in which case \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\) ). $$ \begin{array}{l} \text { Show that } \int \frac{d x}{x \sqrt{x^{2}-1}}= \\ \qquad\left\\{\begin{array}{ll} \sec ^{-1} x+C=\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x>1 \\ -\sec ^{-1} x+C=-\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x<-1 \end{array}\right. \end{array} $$

Use integration by parts to evaluate the following integrals. $$\int_{0}^{\infty} x e^{-x} d x$$

Suppose that the rate at which a company extracts oil is given by \(r(t)=r_{0} e^{-k t},\) where \(r_{0}=10^{7}\) barrels \(/ \mathrm{yr}\) and \(k=0.005 \mathrm{yr}^{-1} .\) Suppose also the estimate of the total oil reserve is \(2 \times 10^{9}\) barrels. If the extraction continues indefinitely, will the reserve be exhausted?

Consider the curve \(y=\ln x\) a. Find the length of the curve from \(x=1\) to \(x=a\) and call it \(L(a) .\) (Hint: The change of variables \(u=\sqrt{x^{2}+1}\) allows evaluation by partial fractions.) b. Graph \(L(a)\) c. As \(a\) increases, \(L(a)\) increases as what power of \(a ?\)
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