the integral of sqrt(9-x^2)/x^2 - math is hot!

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hope this helped! good luck :)
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[@epicpickles9077]

im just starting to learn integrals like this, this was really cool to watch!

[@mint6027]

i recently learned how to do integrals like this... i ❤ math

[@amitgarg804]

can u also post calculus brainrot?

[@mathishot1]

i would argue this entire account is calc brainrot

[@holyshit922]

With integration by parts answer is immediate Integration by parts with u = sqrt(9-x^2) , dv = 1/x^2dx

[@mint6027]

you'll just keep running into integrals forever tho, right?

[@XtronePlaysG]

@@mint6027 if you use integration by parts you will end up with: int of the given = -sqrt(9-x^2)/x + int(1/(sqrt(9-x^2)) which would require trig subbing x = 3sintheta which leads to the same result. So you can use integration by parts but would eventually have to require trig sub anyways, but you wouldn't need to do the triangle identity this way.

[@joserubenalcarazmorinigo9540]

Haciendo por partes se llega, una parte, a la Integral de dx/√(9 - x^2) como indicó correctamente XtronePlaysG, pero aquí ya se puede considerar como Inmediata o sea el resultado es: arc sen (x/3) , por CONOCIMIENTO de Derivadas, o sea ya no es necesario la Sustitución Trigonométrica, y entonces llegaríamos más rápido al resultado. Abrazos

[@holyshit922]

@@XtronePlaysG No if you know derivatives of inverse functions like arcsin you dont need trig sub

[@XtronePlaysG]

@@holyshit922 "if you know the shortcut that skips the trig substitution then you don't need trig substitution" thank you for your valuable input