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I know some integrals can't have undefined integrals, but why? And how, for example, can be proved that the area under the hyperbola $y=\frac{1}{x}$ is $\ln(x)$?

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You have three different questions. The title question is too broad to really answer IMO (different types of integrals have different types of techniques). The second is unclear; are you asking why some functions don't have elementary / closed-form antiderivatives? The third depends on one's definitions of $\ln x$ (sometimes it's defined by the area you mention); if it's the functional inverse $\exp^{-1}x$ then a derivative rule for inverses combined with the fundamental theorem of calculus will do it. – anon Jun 19 '12 at 6:52
I think that "undefined integral" is a translation from a foreign language. In italian we say "integrale indefinito" to mean the set of all primitive functions. – Siminore Jun 19 '12 at 7:36
@Siminore: Yes, we know definite and indefinite integrals. – Babak S. Jun 19 '12 at 8:07
@anon: What are those rules? Post them as an answer, please. – dot dot Jun 19 '12 at 12:03
Beyond merely memorizing the derivatives of elementary functions and then adapting, some broad real-analytic methods are $u$-substitution, by-parts integration, and differentiating under the integral (aka Feynman's trick); sometimes many of these are needed multiple times; sometimes they only work with special definite integrals; sometimes other multivariable methods are needed or useful (eg the Gaussian integral); sometimes substitutions are very tedious (Euler substitutions for elliptic integrals). Complex analysis delivers the very useful tools of contour integration and residue formulas... – anon Jun 19 '12 at 12:13
up vote 2 down vote accepted

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The area under the hyperbola ($y=\dfrac{1}{x}$) between point M and N is $A(x)$ .

The area under the hyperbola ($y=\dfrac{1}{x}$) between point M and L is $A(x+h)$ .

The difference of Area can be defined as $A(x+h)-A(x)$.

If $h\rightarrow0 $ then the difference area will be rectangle thus we can write:

$$ A(x+h)-A(x)\approx \frac{1}{x} h$$

$$ \lim_{h \to 0} \frac{A(x+h)-A(x)}{h}=\frac{1}{x}$$

$$ \lim_{h \to 0} \frac{A(x+h)-A(x)}{h}=\frac{dA(x)}{dx}$$


$$\int dA(x)=\int \frac{dx}{x}$$

$$A(x)=\int \frac{dx}{x}$$




$$\int \frac{dx}{x}=\int dt$$

$$\int \frac{dx}{x}=t$$

$$\ln x=\ln e^t=t \ln e=t$$

$$\int \frac{dx}{x}=\ln x$$

$$A(x)=\int \frac{dx}{x}=\ln x$$

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Every function that is finitely discontinuous has an indefinite integral. The problem is: Some integrals can't be written in the form of elementary functions.

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Some is more likely most. – lhf Jun 19 '12 at 13:49
i agree with you lhf. – Aang Jun 20 '12 at 5:27

You must use the fundamental theorem of calculus, essentially that differentiation and integration are inverse to each other.

It can be shown that the derivative of ln$(x)$ wrt $x$ is $\frac{1}{x}$ and so ln$(x)$ is a suitable function to be taken for the integral of $\frac{1}{x}$ upto addition of a constant.

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Oops :p edited. – fretty Jun 19 '12 at 12:15

If the function being integrated has singularities, the integral can only be defined by breaking the integral up into limits approaching the singularity from whichever sides are appropriate. So the reason it is fine for some functions and not fine for others has to do with whether or not the limit of the integral as you approach the singularity exists or not.

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