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Integration:
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\[\LARGE \int~\frac{dx}{\sqrt{x^2-6x+13}}~dx\]
complete the square for the quadratic inside bottom radical, then make a trig substitution
So something like? \[\LARGE \int~\frac{dx}{\sqrt{(x-3)^2+4}}~\] Then sub \(u=x-3\) and go from there>
Yep ! or make the substitution directly in single step : sub \(\large x-3 = 2\tan u\)
Wait, *-4, not 4.
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+4 is right : 9 + 4 = 13
Oh, right, sorry, did that too quickly. Btw, isn't \(\Large \int~\frac{dx}{\sqrt{1+x^2}}=sinh^{-1}x\) not tan?
that works too^
Alright, thank you rational :)
earlier substitution is for ppl who are familiar wid hyperbolics yet, yw :)
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