Integrals :)
\[\int\limits 8(x+x^2)^{-1}~dx\]
Would I distribute the 8 first?
\[\int\limits_{ }^{ } \frac{8}{x(1+x^2)} dx \]
I would just move the 8 out of the integral :) It's just a constant
wouldn't that be x(1+x) on the bottem nnesha
Whoops, that was supposed to be 1+x^2
\(\displaystyle \int \frac{8}{x^2+1} dx\). Trig sub.
or just recon the derivative of what may the integrand be?
(You have done the derivatives of inverse trig functions, haven't ya?)
Yes, we've done inverse sine, tangent and secant
Yes ... more is not requered.
Oohhh, I think I see, it would be 8\(tan^{-1}x + c\)?
Do you remember the inverse derivative of \(\arcsin(x)\), \({\rm arcsec}(x)\), and \(\arctan(x)\)?
did I just say INVERSE DERIVATIVE ?? OMG ... Just derivative (not inverse derivative)
yes, you are thinking correctly:)
We haven't done any of those 3...
Though, if you want to actually solve it, as opposed to realizing the derivative, you may choose to ask for more assistance. (If you learned trig sub, I would advise that, if not, don't worry right now, you will cover it later.)
We haven't gotten into that yet, just started integrals today :)
Oh today ... you are getting tuned in very quickly then. Keep it up!
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