# How integrals are computed?

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 $\ln(x)$?

-
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. –  B. 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

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:

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

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

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

$$\frac{dA(x)}{dx}=\frac{1}{x}$$

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

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

$$x=e^t=1+\frac{t}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\cdots$$

$$\frac{dx}{dt}=e^t=1+\frac{t}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\cdots=x$$

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

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

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

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

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

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

-

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.

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