I want to integrate a difficult single variable function but I don't want to modify the given function i.e. factorising, rationalizing, etc. The question is: $$f(x)= (x^3-4x^2+4x)(x^2-4)^{-1}$$
The answer turns out to be equals to: $g(x)= (x^2/2)+4\ln(|x-4|)+C$ after factorising the question. But I don't want this! If you can see, there is something special about the question itself. At $x=2$ the $f(x)=0/0$. But when we calculate derivative of $g(x)$ we get, $h(x)= (x^2-4x+4)(x-4)^{-1}$ but $h(x)$ is not equals to $0/0$ at $x=2$... So I want a $F(X)$ whose derivative will give "exactly" my original given $f(x)$- no factorization, no rationalization, etc... (You can use u-substitution but be careful my objective shall be achieved) and I'm well aware that $h(x)$ is equivalent to $g(x)$... In short, I want a F(X) such that its derivative is exactly equals to my given f(x)- it doesn't matter what techniques you use! You can even use technique other than integration if possible.
Thank you for spending your precious time Respected Regards, Swayam Jha
The trouble is that you are treating the indeterminate form $f(2)=0/0$ as a number, while $f(2)$ is well defined, as you can verify by looking at the graph. Applying L'Hopital (or by whatever method you like$^\dagger$), we see that $f(2)=0$. As expected, $g'(2)=0$, so $g'(2)=f(2)$. If you really wanted, I suppose you could write
$$ g'(x)=\frac{4}{x-4}+x=\frac{x^2-4x+4}{x-4}$$
Multiply top and bottom by $x-2$
$$ g'(x)=\frac{(x^2-4x+4)(x-2)}{(x-4)(x-2)}=\frac{x^3-6x^2+12x-8}{x^2-6x+8}$$
Which also has the indeterminate form $0/0$ at $x=2$
EDIT: actually, just multiply $g'$ by $1=x/x$ and we recover your original function
$$ g'(x)=\frac{(x^2-4x+4)(x)}{(x-4)(x)}=\frac{x^3-4x^2+4x}{x^2-4x}=f(x)$$
$\dagger$ It is simplest by reversing the last step (dividing by $1=x/x$)
$$f(x)= \frac{x^3-4x^2+4x}{x^2-4x}=\frac{x^2-4x+4}{x-4}$$
Evaluating the right hand side at $x=2$ gives $0/2=0$