Question about "Expanding the integral."
so I'm curious how we get things like 1/(sqrt(x^4) = sqrt(x^4)/x^4
or that sqrt(x^4)/x^4 = 1/x^2
Is this one approach to many for doing a question like this? Sometimes Wolfram does some "interesting" things...
i dont know about the integral part, but sqrt(x^4)/x^4 = 1/x^2 because the square root of x^4 is x^2. and then the bottom can become x^2. so its 1/x^2
Ty Jessica.
are you familiar with the magic \[\frac{1}{\sqrt x} \implies \frac{1}{\sqrt x} \times \frac{\sqrt x}{\sqrt x} \implies \frac{\sqrt x}{x}\]
it follows the same principle
no.
you're not familiar with rationalization?
teach me about everything.
i find that very surprising...
hmm now that I look at that I think I've seen that before... So we do this with sqrt's usually?
rationalization is done when you have a radical in the denominator..your goal is to "neutralize" the denominator and depends on the situation really
hmm cool trick :)
I have been denied this cool tricks my entire childhood.
for example you have \[\int \frac{x-1}{\sqrt x}dx\] there's no need for rationalization...just split up the integral \[\int \frac{x}{\sqrt x} - \int \frac{1}{\sqrt x}\] you can use rationalization but you know this is just \[\int x^{1 - 1/2} - \int x^{-1/2}\]
\[x^{1 - 1/2} = x^{1/2}\] now notice if i use rationalization \[\frac{x}{\sqrt x} \implies \frac{x}{\sqrt x} \times \frac{\sqrt x}{\sqrt x} \implies \frac{x\sqrt x}{x} \implies \sqrt x \implies x^{1/2}\]
both will result the same answers
so if you had a x^1/2 you could randomly do x/(x)^(1/2)
you're soo cool iggy.
TEACH ME MOAR.
@lgbasallote run away after medal eh?
lol i ran away much earlier :p
and why would you make x^1/2 into a more complicated state?
i guess if you have some outside of the box plan...it's possible
I'm getting this from another calculator
I'm sure there's something to do with "restricted values of x" in wolfram
wolfram
:)
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