Can you multiply an integral by $f(x)/f(x)$ where $deg(f(x))>0$?

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I was on instagram and someone posted this: https://www.instagram.com/p/BplPOrlAvob/

I haven't been on here in some time, so please excuse my poor formatting, but the integral to be solved is:

$\int {x^4 \over x^2 +1}dx$

In his solution, he begins by multiplying the function by ${x^2-1} \over {x^2-1}$

I seem to remember my professor in first year saying you should not multiply by functions because it changes the integral. I was just wondering if I am just imagining this or if I am remembering correctly.

It has also been around 3-4 years since I've done analysis but I would really appreciate a mathematical explanation if it is the case. Is it because multiplying by a function creates a hole? Or does a problem only arise if you were to integrate over discontinuous intervals?

Thank you!

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Multiplying by $\frac{x^2-1}{x^2-1}$ is perfectly valid. Except for removable discontinuities this evaluates to 1, so it does not change the integrand and thus the integral.

However, for this integral such multiplication is redundant; we can straight away write $$\frac{x^4}{x^2+1}=\frac{x^4-1+1}{x^2+1}=x^2-1+\frac1{x^2+1}$$ and thus skip to the second-last line in the Instagram post. I suppose those extra steps were made to "look smart".