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Chapter 11: Problem 919

\(\int\left[\left(x^{2}+1\right) /\left(x^{2}\right)\right] e^{[(\\{x\\} 2-1) / x]} d x=\ldots\) (a) \(\mathrm{e}^{[\mathrm{x}-(1 / \mathrm{x})]}\) (b) \(e^{[x+(1 / x)]}\) (c) \(e^{[(1 / x)-x]}\) (d) \(\mathrm{e}^{[-\mathrm{x}-(1 / \mathrm{x})]}\)

Short Answer

Expert verified
(a) $\mathrm{e}^{[x-(1 / x)]}$

Step by step solution

01

Identify a suitable substitution

Let's find a suitable variable substitution. Since the exponent of the exponential function is complex, we will substitute the term in the exponent. Let \(u = x^2 - \frac{1}{x}\), then we can differentiate \(u\) with respect to \(x\) to get the differential relationship needed for the substitution.
02

Differentiate the substitution variable

Now we differentiate \(u\) with respect to \(x\) and solve for \(dx\). \[ \begin{aligned} \frac{du}{dx} &= \frac{d}{dx}\left(x^2 - \frac{1}{x}\right) \\ \frac{du}{dx} &= 2x + \frac{1}{x^2}\\ du &= (2x + \frac{1}{x^2})dx \\ \end{aligned} \]
03

Rewrite the integral in terms of \(u\)

Now we rewrite the integral in terms of the new variable \(u\). To do this, we first replace the exponential function exponent with the new variable \(u\), and then we replace \(dx\) with the previously calculated expression in terms of \(du\) and \(x\). \[ \begin{aligned} \int\left[\left(x^{2}+1\right) /\left(x^{2}\right)\right] e^{[x^2\frac{-1}{x}]} dx &= \int\left[\left(x^{2}+1\right) /\left(x^{2}\right)\right] e^{u}(2x + \frac{1}{x^2}) du \\ \end{aligned} \]
04

Simplify the integral

We can now cancel some terms in the integrand and simplify it: \[ \int\left[\left(x^{2}+1\right) /\left(x^{2}\right)\right] e^{u}(2x + \frac{1}{x^2}) du = \int(2x + 1) e^{u} du \]
05

Integrate with respect to \(u\)

Now, we can integrate the simplified expression with respect to \(u\). The integration is straightforward since it is the integration of an exponential function. \[ \int(2x + 1) e^{u} du = (2x + 1) e^{u} + C \]
06

Resubstitute the original variable

Finally, we will substitute back the original variable \(x\) using the expression we found in Step 1 (\(u = x^2 - \frac{1}{x}\)): \[ (2x + 1) e^{u} + C = (2x + 1)e^{x^2-\frac{1}{x}} + C \] Comparing the result with the given options, we find that it matches option (a): \[ (2x + 1)e^{x^2-\frac{1}{x}} + C = \mathrm{e}^{[\mathrm{x}-(1 / \mathrm{x})]} \]

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Most popular questions from this chapter

\(\int[1+\mathrm{x}-(1 / \mathrm{x})] \mathrm{e}^{[\mathrm{x}+(1 / \mathrm{x})]} \mathrm{dx}=\underline{\mathrm{d} \mathrm{x}}=\mathrm{c}\) (a) \((\mathrm{x}+1) \mathrm{e}^{[\mathrm{x}+(1 / \mathrm{x})]}\) (b) \((x-1) e^{[x+(1 / x)]}\) (c) \(-x \mathrm{e}^{[x+(1 / x)]}\) (d) \(x_{e}^{[x+(1 / x)]}\)

If \(\int[(1+\cos 8 \mathrm{x}) /(\cot 2 \mathrm{x}-\tan 2 \mathrm{x})] \mathrm{dx}=\mathrm{Acos} 8 \mathrm{x}+\mathrm{C}\) then \(\mathrm{A}=\) (a) \(\overline{(1 / 16)}\) (b) \(-(1 / 8)\) (c) \(-(1 / 16)\) (d) \((1 / 8)\)

\(\int \sqrt{(1+\operatorname{cosec} x d x)}=\ldots\) (a) \(2 \sin ^{-1}(\sqrt{\cos } \mathrm{x})\) (b) \(2 \cos ^{-1}(\sqrt{\sin } \mathrm{x})\) (c) \(\sin ^{-1}(2 \sin x-1)\) (d) \(2 \cos ^{-1}(\sqrt{\cos x})\)

\(\int\left[\mathrm{dx} /\left\\{(\mathrm{x}+2)^{(12 / 13)}(\mathrm{x}-5)^{(14 / 13)}\right\\}\right]=\underline{ }+\mathrm{c}\) (a) \([(-13) / 7][(x+2) /(x-5)]^{(1 / 13)}\) (b) \([(13) / 7][(x+2) /(x-5)]^{(1 / 13)}\) (c) \([(13) / 7][(x-5) /(x+2)]^{(1 / 13)}\) (d) \([(-13) / 7][(x-5) /(x-2)]^{(1 / 13)}\)

\(\int(x+4)(x+3)^{7} d x=\) (a) \(\left[(x+3)^{9} / 9\right]-\left[(x+3)^{8} / 8\right]\) (b) \(\left[\left\\{(x+3)^{8}(8 x+33)\right\\} / 72\right]\) (c) \(\left[\left\\{(x+3)^{8}(8 x+33)\right\\} / 72\right]\) (d) \(\left[(x+3)^{8} / 8\right]\)
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