I started with the integral of:
$$\int \frac{\ln(x)}{x}dx = \frac{\ln^2(x)}{2} + C$$
Which was very easy to integrate. Then, I moved to a more difficult problem: $$\int \frac{\ln(x-t)}{x} dx$$
I started by letting $t$ tend to some arbitrary, zero-like variable $O_t$. Where some $a_0 - O_t = a_0$, but $O_t$ is not technically equal to zero. In other words, expressions containing $O_t$ cannot be simplified and compacted, but it behaves like zero when the cyclical nature of the integral pops up.
Using it, the DI method yields: $$\int \frac{\ln(x-O_t)}{x}dx = \ln(x-O_t)\ln(x) - \int \frac{\ln(x)}{x-O_t}dx$$
Since this $O_t$ is technically equal to zero, then by definition: $$\int \frac{\ln(x-O_t)}{x}dx = \int \frac{\ln(x)}{x-O_t}dx$$
Plugging this is, you find that: $$\int \frac{\ln(x-O_t)}{x}dx = \ln(x-O_t)\ln(x) - \int \frac{\ln(x-O_t)}{x}dx$$
You can add the integral to both sides, and divide by two: $$\int \frac{\ln(x-O_t)}{x}dx = \frac{\ln(x-O_t)\ln(x)}{2}+C$$
Then, in order to complete the problem, you can let $O_t$ re-approach $t$:
$$\int \frac{\ln(x-t)}{x}dx = \lim\int \frac{\ln(x-O_t)}{x}dx = \frac{\ln(x-t)\ln(x)}{2} + C$$
In order to test this, you can plug in $t=0$ to see this conclusion in action: $$\int \frac{\ln(x-0)}{x} dx = \frac{\ln(x-0)\ln(x)}{2} + C = \frac{\ln^2(x)}{2} + C$$
Which indeed is correct. Is this new "zero-substitution" technique for integration viable? Or does it have major gap holes. Should I pursue it and apply it to other integrals?
Thank you for your help.
It is not true that $$\int \frac{\ln(x-t)}{x} \ dx = \frac{\ln(x-t)\ln(x)}{2} +C$$ in general because differentiating the right hand side we obtain $$\frac{1}{2}\left(\frac{\ln(x-t)}{x}+\frac{\ln(x)}{x-t}\right).$$ In the case where $t=0$ this does hold, but also if $t=0$ then $$\int \frac{\ln(x)}{x} \ dx = \ln(x)^{2} - \int \frac{\ln(x)}{x} \ dx$$ and the result follows similarly as in your post without the need for $O_{t}$.