Question:
Find the following integral: $\int{(1+x^{-1}+x^{-2})dx}$
My book's attempt:
$$\int{(1+x^{-1}+x^{-2})dx}$$
$$...$$
$$x+\ln x-x^{-1}\tag{1}$$
Shouldn't the answer be $x+\ln|x|-x^{-1}$ instead of $(1)$?
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7The answer should be $x + \ln |x| - x^{-1} + C$, where $C$ is a constant. – N. F. Taussig Jan 24 '22 at 14:45
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3Does this answer your question? What is the integral of 1/x? – Hermis14 Jan 24 '22 at 14:46
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2Yes the answer should be $x + \ln |x| - x^{-1} + C$ with $C$ the constant of integration. – Essaidi Jan 24 '22 at 14:46
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3Yes. But is there other context in the problem that would let you assume $x>0$? – electronpusher Jan 24 '22 at 14:46
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7If we really want to describe all anti-derivatives on the largest possible domain, the most general answer is actually $$ \begin{cases} x + \ln(-x) - x^{-1} + C_1 & x<0 \ x + \ln x - x^{-1} + C_2 & x>0 \end{cases} $$ – aschepler Jan 24 '22 at 14:52
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@electronpusher no sir, no context is given that would let me assume that x>0. – tryingtobeastoic Jan 24 '22 at 15:06
1 Answers
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This is debatable.
If you want to apply the FTC, you will subtract two values of this antiderivative,
$$\int_a^b\frac{dx}{x}=\log b-\log a=\log\frac ba$$ or $$\int_a^b\frac{dx}{x}=\log|b|-\log|a|=\log\left|\frac ba\right|.$$
But as the integrand has a singularity at $x=0$, $a,b$ may not differ in sign.
For this reason, I prefer the non-standard expression.
Also note that
$$\begin{cases}x<0\to\log(-x)+C_-,\\x>0\to\log(x)+C_+\end{cases}$$ is also a valid antiderivative.
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2But the first version doesn't easily describe $\int_{-2}^{-1} \frac{dx}{x}$, without the complication of a complex log branch. – aschepler Jan 24 '22 at 14:55
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1Why would the comments and textbooks know what you are trying to say? I was merely suggesting a disambiguation edit. – ryang Jan 24 '22 at 15:07
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@aschepler Untrue, since you can use the antiderivative family you yourself provided, which coincides with Bobby's, to apply the fundamental theorem of calculus: refer to the part where $x\lt0.$ – Angel Jan 24 '22 at 20:23