I know that the integral of$1/x^2$ is $(-1/x +C)$
Can we split $(1/x^2)$ as $(1/x)(1/x)$ and say that the integral of $(1/x^2)$ is $\ln x\cdot\ln x$?
PS: I am new to high school. And this is my first question on Stack Exchange.
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$\begingroup$It is good to ask no problem. The integral behaves nicely with addition and subtraction , I mean, if $\int f(x) dx$ and $\int g(x) dx$ exist then $$\int(f(x)+g(x) dx=\int f(x) dx+\int g(x) dx$$ The same for subtraction. But in general , we can NOT say $$\int f(x) g(x) dx=\int f(x) dx \times \int g(x) dx$$I hope this helps you.
$\endgroup$ 2 $\begingroup$If you know how to differentiate a product you will know that $(fg)'=f'g+fg'$ (where $f'$ is the derivative of $f$). Set $h=f'$ so that $f=\int h dx $.
Then $(fg)'=hg+fg'$ and $$\int hg dx=\int\left((fg)'-fg'\right)dx=fg-\int(fg')dx$$which is the formula for integration by parts.
Because I began from the derivative of a product, the notation for integrals gets a bit clunky. zwim in the comments has usefully given a more convenient form, which simply comes from renaming the functions, with $f=F'$:
$$\int fg dx=Fg-\int Fg' dx$$
The formula for the derivative of a product is one of its characteristic properties, which gets generalised to derivatives in other contexts. Its form suggests that integration, as the inverse operation, will have a more complicated relationship with the product of functions than you have postulated. But "how does this work with products" - what is the relationship of some operation with the underlying arithmetic - is always a good one to ask.
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