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 $\ln(x)$?
 A: 
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:
$$ A(x+h)-A(x)\approx \frac{1}{x} h$$
$$ \lim_{h \to 0} \frac{A(x+h)-A(x)}{h}=\frac{1}{x}$$
$$ \lim_{h \to 0} \frac{A(x+h)-A(x)}{h}=\frac{dA(x)}{dx}$$
$$\frac{dA(x)}{dx}=\frac{1}{x}$$
$$\int dA(x)=\int \frac{dx}{x}$$
$$A(x)=\int \frac{dx}{x}$$
$$x=e^t=1+\frac{t}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\cdots$$
$$\frac{dx}{dt}=e^t=1+\frac{t}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\cdots=x$$  
$$\frac{dx}{dt}=x$$
$$\int \frac{dx}{x}=\int dt$$
$$\int \frac{dx}{x}=t$$
$$\ln x=\ln e^t=t \ln e=t$$
$$\int \frac{dx}{x}=\ln x$$
$$A(x)=\int \frac{dx}{x}=\ln x$$
A: 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. 
A: 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.
A: 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.
