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Never used notion before, I used to use obsidian

Factor out a negative and flip the denominator (this is like multiplying by two negatives and the fire has no effect on the overall value) to where it’s now -1 times the integral of (e^x + 1)/(e^x - 1) now let u = e^x - 1 so du = e^xdx the integral becomes -1 times integral of (u + 2)/(u^2 + u) = -1/2 times integral of (2u + 4)/(u^2 + u) = (2u + 1 + 3)/(u^2 + u) = -1/2 times integral of (2u + 1)/(u^2 + u) - 3/2 times integral of 1/u(u+1). Perform partial fraction on second integral and evaluate both and you get (-1/2)ln|u^2 + u| - (3/2)ln|u| + (3/2)ln|u+1| Apply rules for logarithms to simplify (-1/2)ln|u^2 + u| = (-1/2)ln|u| - (1/2)ln|u+1| whole expression becomes ln|u + 1| - 2ln|u| Now sub back in for x ln|e^x| - 2ln|e^x - 1| + C = x - 2ln|e^x - 1| + C

Let dv = xdx and v = (x² + 1)/2

i multiplied and divide e^x/2 and then took e^-x/2 - e^x/2= t, the soln i am getting is -2ln(e^x/2-e^x/2) + c , is it correct

nyc explanation ❤ but one suggestion - if you write code in video , it will be highly encouraging.

Cant we do substitution ? 1+x^2 = t , then using byparts ?

Always love a good partial fraction. I used to love doing these as stress relief.

I’m curious, if I was in the actual MIT integration bee, would it be incorrect if I didn’t give them this exact solution? Technically, if I didn’t combine the two constants at the end, I’d still have a valid antiderivative to my original integral. Would they still accept it?

Simplă, dar educativă. Așa se pornește de la simplu la complex. Cine vrea, poate să învețe. Succes în continuare. Simplă pentru cine știe, pentru cine nu știe, e grea. Iubesc matematica la infinit !

Integrale nedefinite de fincții(primitive de funcții) relativ ușoare, frumoase, bine alese, antrenament pentru rezolvarea altora cu grad de dificultate mai mare. Felicitări, vă doresc succes în continuare. Iubesc matematica la infinit !!!!!!

Wathc next (at the end) gives empty box

So nice, thanks

Thanks bro Mohammed Ma sha Allah

Thanks it was so nice

I did it a bit differently. Let x/root(2) = sec(y) dx = root(2)sec(y)tan(y)dy 1/root(2) times integral of 1 dy (trig identities simplify everything down to 1 alongside the substitution) and you get (1/root(2))sec^-1(x/root(2)) + C EDIT if tan(y) = root((x^2 - 2)/2) then opposite side = root(x^2 - 2) and adjacent side is root(2) so hypotenuse is x therefore tan^-1((root(x^2 - 2)/root(2)) = sec^-1(x/root(2)) Our answers are indeed equivalent.

bro i have come back again i got selected into isi bmath you remembered me??

Sorry for the noise :)

nice integral , i did it by substituting 1/x=t

∫(x²+1)/(x+1)dx = ∫(x²-1+2)/(x+1)dx = ∫(x²-1)/(x+1)dx + ∫2/(x+1)dx and x²-1 = (x-1)(x+1) = ∫(x-1)dx + 2∫1/(x+1)dx = x²/2 - x + 2ln|x+1| + c

Animations are so smooth. I would also like to learn how to make such videos. Can you please tell me which software do you use?

same thing. I just combined the final two logarithms into (1/4)ln|(x^2 + 1)/(x^2 - 1)| + C

I don't get it

Bro, put your video code in your description, this may help some people

After the first step to seperate x^1 from the integral everything became pretty trivial

How do I learn manim professionally?

I think you should do more videos explaining it on your tablet, it really allows your personality to shine through

3Cos(3x + 5 ) is the answer

Now Im back too 😎😎😎😎(I am finally not sick). Also can you assume the principal square root to be positive in a definite integral like we did here?

Let's go gambling!

Thats a good one

Very nice explanation sir and we understand logic

poor explanation, no tree diagram, no reasoning wastage of time

sound quality is bit lacking...

what about suffix functionality

great explanation :)

Anyways, if you're going through the comment section, a like would be really appreciated :)

Please speak little loud sir , your explaination is nice but we have to make effort to hear you

The volume is quite low.

Solution: ∫(x²+1)/(x+1)*dx = --------------- Substitution: u = x+1 x = u-1 du = dx --------------- = ∫[(u-1)²+1]/u*du = ∫[(u²-2u+1)+1]/u*du = ∫(u²-2u+2)/u*du = ∫(u-2+2/u)*du = u²/2-2u+2*ln|u|+C = (x+1)²/2-2*(x+1)+2*ln|x+1|+C = (x²+2x+1)/2-2x-2+2*ln|x+1|+C = x²/2+x+1/2-2x-2+2*ln|x+1|+C = x²/2-x-3/2+2*ln|x+1|+C = x²/2-x+2*ln|x+1|+C-3/2 = x²/2-x+2*ln|x+1|+D Checking the result by deriving: [x²/2-x+2*ln|x+1|+D]’ = x-1+2/(x+1) = [x*(x+1)-1*(x+1)+2]/(x+1) = [x²+x-x-1+2]/(x+1) = (x²+1)/(x+1) everything okay!

I think it’s ln|sech(x)|+C

Its -1 without watching the video

I wanna watch all of these but I'm so sick 😭😭😭😭

Take lnx common and write 2+lnx as 1+1+lnx then multiply in lnx back. Integrate lnx and lnx(1+lnx) separately. Note that 1+lnx is the differentiation of xlnx so we can apply by parts here. This will be easier.

Add -1 and +1 its easier

how to set my reels and shorts feed to only things that will immediately benefit me?

Yes, this is quite helpful for beginners who want to learn programming from scratch. Next topic should be on.. operations on data types. Right? For example, string concatenation, updating variables like x=x+1 and so on.

Okay.

Why wait? Just do it perfectly the first time.

Why did xlnx = lnt turn to (x*1/x+lnx)dx = 1/t dt