Integration of two variable function

Note: This concept idea hasn't been proven yet or maybe mathematically incorrect. Any contribution would be greatly appreciated.

Basic Knowledge of Integration and Functions

In mathematics, usually we take x as independent variable and y as dependent variable. In General, a function is written as

y=ax^2+bx+c

If we differentiate the above function, we get:

\frac{dy}{dx}=2ax+b

i.e  dy=(2ax+b)dx

Generally, a Integral function is written as:

f(x)=\int (2ax+b) dx

Note: There is a wrong perception that, Integration give area under function, but actually, Definite Integration give area under a given function f(x) and Indefinite Integration gives an entire family of functions with an infinite number of members.

Therefore, it is understood that:

f(x)= \int dy=\int (2ax+b)dx

So, After clearing the basic knowledge, we can see how to Integrate something like this:

\int f(x,y) dx   (where both x and y are variable)

Integration of two variable function F(x,y)

Lets take an example function:

f'(x,y) = 3x^2+y

So, how to find: \int (3x^2+y)dx

As mentioned above, it is understood that it means:

\frac{dy}{dx}=3x^2+y   (and we have to find its integral)

\implies \frac{dy}{dx}-y=3x^2   (Rearranging the Equation)

Now we have convert the Integral Question, into a Differential Equation.

So, we can solve this using, P,Q Factor method (What is Factor Method ?)

\frac{dy}{dx}-y=3x^2

Let P=(-1) and Q=3x^2

Therefore:

Integral Factor (I.F)=e^{\intP dx}

\implies (I.F) = e^{\int-1 dx}

\implies (I.F) = e^{-x}

Hence our result is:

y(I.F)=\int Q (I.F) dx

\implies ye^{-x}=\int 3x^2e^{-x} dx

\implies ye^{-x} =-3(x^2e^-x +2xe^-x +2e^-x) +C   (using By Parts)

\implies ye^(-x)+3x^2e^{-x}+6xe^{-x}+6e^{-x}=C   (rearranging the equation)

\implies \ln(e^{-x}(y+3x^2+6x+6))=\ln C    (taking log both the sides)

\implies -x+\ln(y+3x^2+6x+6)=A

Hence, f(x,y) =  -x+\ln(y+3x^2+6x+6)=A

Fun Assignment: Differentiate the Above function, and you will get the original f'(x,y)