26:00 i think it's called the cover up method where you basically let x +1 equals zero. X = -1, so you cover up the x+1 and put -1 in the original equation. That value is the numerator of x+1. And then you can find b and c.
complexifying the integral is a technique you're talking about lol, I can't do that xD i like to do things basic-ish, idk anything about complex analysis or stuff like that
Silver i mean, the basic thing about this is knowing real and imaginary numbers. It's somewhat easy once you get used to it, and again i've seen some people used it in integration bee. My only reason of using this is just this saves so much time. I could give a bit of background on the definition. If you're not interested, that's okay. Again the only reason this is viable is it saves time.
Silver the definition starts by putting the imaginary number on the number exponential e. Where imaginary number i is sqrt of -1. Consequently it's exponents is imaginary, then when you look at the taylor series, you have something that alternates sign every 2 terms. This is similar to our cosine and sine taylor series. Multiplying the sine taylor series by i and adding the cosine series gives you e^i. So there's two parts. The real part which is the cosine and the imaginary part which is the sine. The only important thing is if something has the sinx function and you want to integrate, you can just replace the sine with e^ix and treat i as a constant. Of course the next thing you have to know is to just play with complex numbers. If the denominator has i in it, multiply top and bottom by its conjugate. And the final value has two parts, the imaginary and the real part. You just look at anything that has i in it, and that's the antiderivative of our original function, removing the i.
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26:00 i think it's called the cover up method where you basically let x +1 equals zero. X = -1, so you cover up the x+1 and put -1 in the original equation. That value is the numerator of x+1. And then you can find b and c.
I believe the answer would be 2/3 ln(x+1) + 1/6 ln(x^2-x+1) + 1/(sqrt3) arctan((2x-1)/sqrt3)
8:00 if you know complex numbers then you know what to do
I've seen some experienced integrator used it
complexifying the integral is a technique you're talking about lol, I can't do that xD
i like to do things basic-ish, idk anything about complex analysis or stuff like that
Silver i mean, the basic thing about this is knowing real and imaginary numbers. It's somewhat easy once you get used to it, and again i've seen some people used it in integration bee. My only reason of using this is just this saves so much time. I could give a bit of background on the definition. If you're not interested, that's okay. Again the only reason this is viable is it saves time.
Silver the definition starts by putting the imaginary number on the number exponential e. Where imaginary number i is sqrt of -1. Consequently it's exponents is imaginary, then when you look at the taylor series, you have something that alternates sign every 2 terms. This is similar to our cosine and sine taylor series. Multiplying the sine taylor series by i and adding the cosine series gives you e^i. So there's two parts. The real part which is the cosine and the imaginary part which is the sine.
The only important thing is if something has the sinx function and you want to integrate, you can just replace the sine with e^ix and treat i as a constant. Of course the next thing you have to know is to just play with complex numbers. If the denominator has i in it, multiply top and bottom by its conjugate. And the final value has two parts, the imaginary and the real part. You just look at anything that has i in it, and that's the antiderivative of our original function, removing the i.
Same thing with cosine but, instead we take the real part.