how about if you can generalize this. i always try and generalize the integrals. so we have
integral (c+dx) / sqrt ( ax^2+bx+c)
Now using your idea we multiply top and bottom by 2a/d , which gives us
d/ (2a) * integral ( 2ac/d + 2ax ) / ( sqrt ax^2 + bx + c) . Now on the top we add and subtract by b. so we have d/(2a) integral ( 2ax + b +2ac/d - b ) / sqrt (ax^2 + bx + c).
sorry a lot of letters. now you have d/(2a) integral u^-1/2 du + [(2ac)/d - b] integral 1/sqrt ax^2 + bx + c
his answer is wrong. in the second integration U must be 2X not X. check wolfram (the online integral solver, just google it) and the answer is different.
this one was kind of difficult with all these things in the last ppart..i got mixed up when i tried it =( nice job anyway this guy explains very very well
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LeconsdAnalyse 5 months ago
did you forget the 1 over root 2 ? lol
hq319 6 months ago
im scared of collage now
kanopopo 1 year ago
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how about if you can generalize this. i always try and generalize the integrals. so we have
integral (c+dx) / sqrt ( ax^2+bx+c)
Now using your idea we multiply top and bottom by 2a/d , which gives us
d/ (2a) * integral ( 2ac/d + 2ax ) / ( sqrt ax^2 + bx + c) . Now on the top we add and subtract by b. so we have d/(2a) integral ( 2ax + b +2ac/d - b ) / sqrt (ax^2 + bx + c).
sorry a lot of letters. now you have d/(2a) integral u^-1/2 du + [(2ac)/d - b] integral 1/sqrt ax^2 + bx + c
markov2b1 1 year ago
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markov2b1 1 year ago
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markov2b1 1 year ago
very clever. great video!!!
markov2b1 1 year ago
how do you solve the integral 1/(sqrt(25-x^2))dx
goamira 1 year ago
@goamira
first factor out the 25 from the inside the square root. this gives you 1/5 * integral ( 1 / sqrt (1- (x/5)^2)) , now u = x/5
so 5du = dx, so the 1/5 and 5 cancel. you are left with integral 1 / sqrt ( 1 - u^2) = arcsin u + c , or arcsin x/5 + c
markov2b1 1 year ago
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mtrem2212 1 year ago
Kool thank you!
ImmanualKant1989 1 year ago
why can't the end just be ln of the entire bottom times its derivative? since it's one over the denominator.
ksonggg 1 year ago
@ksonggg Because that's not the actual rule. You're thinking of just ln(x).
But the derivative of ln[f(x)] = f'(x)/f(x). e.g. f(x) = x^2
ln(x^2) => 2x/x^2 and so on.
ElliotTM92 1 month ago
:D :D :D thanks!!!!!!
tamaratm22 1 year ago
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LeconsdAnalyse 1 year ago
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LeconsdAnalyse 1 year ago
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LeconsdAnalyse 1 year ago
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LeconsdAnalyse 1 year ago
his answer is wrong. in the second integration U must be 2X not X. check wolfram (the online integral solver, just google it) and the answer is different.
sheazer 2 years ago
Thanks. This was very helpful.
Keep making these videos, please.
Coundn't you do something on series and sequences, please?
lcunha85 3 years ago
You forgot two things:
The 1/rt(2) on the numerator in the second half of the video and the du/rt(u) du.
Other than that, I found this video VERY useful and it's an excellent example.
gamesguru 4 years ago
lol tome is cute cuz he got jokes. holler back at your man bro.
bkjoelover 4 years ago 2
Ugh... damn integral calculus.
259 4 years ago
the one over root 2 doesnt really matter...it's just a constant and can be carried out of the integral (put it in front)
tingcong222 4 years ago
you missed the 'one over square root of 2' in line three (check video @ time 7:25)
newazninvasion 4 years ago 2
nasty problem
jonathancolucci 4 years ago
you missed the 'one over square root of 2' in line three (check video @ time 7:25)
newazninvasion 4 years ago
butt nasty alrite
bkjoelover 4 years ago
some good stuff, all be it abit rushed towards the end.
hence the mistake of omitting the sqrt of 2 & also talking about u to the power of 3/2 instead of 1/2 in the 1st integrand.
i'd also like to see them confirm the integral by differentiation. they may then show us some differentiation tricks.
19madrid42 4 years ago
tom is a bad boy
bkjoelover 4 years ago
this one was kind of difficult with all these things in the last ppart..i got mixed up when i tried it =( nice job anyway this guy explains very very well
hecatonchiress 4 years ago
Very nice, thanks.
chrisrave 5 years ago