The roots of the denominator are just the 4th roots of -1, which we can group into complex conjugate pairs (r1, r2) = e^(±pi*i/4) and (r3, r4) = e^(±3*pi*i/4). (x-r1)*(x-r2) = x^2 + 2*cos(pi/4)*x + 1 = x^2 + √2*x + 1 (x-r3)*(x-r4) = x^2 + 2*cos(3*pi/4)*x + 1 = x^2 - √2*x + 1 therefore, x^4 + 1 = (x^2 + √2*x + 1)*(x^2 - √2*x + 1) Also, if you don't mind having logarithms with complex arguments and coefficients in the final result, the problem becomes very straightforward. The result being r/4 * (ln((x+r)/(x-r)) + i*ln((x+i*r)/(x-i*r))) where r = e^(pi*i/4)
Divide both numerator and denominator by x^2. Then by algebraic manipulations, we can write the ratio as sum of two ratios which can be integrated individually.
The process of manipulation of numerators can be made tidier by using polynomial division: divide numerator with the derivative of the denominator polynomial and then i t naturally splits.
Podemos reescribir: 1/(x^4+1)=[1/2*(x^2+1-√2 x)+1/2*(x^2+1+√2 x)]-x^2]/(x^4+1) Luego, al integrar, separamos los 3 sumandos. Notaremos que las 2 primeras integrales son directas: I1=int 1/2*1/[(x+√2 x)^2+(√2/2)^2] -> arctan Igualmente, I2 -> arctan I3=int x^2/(x^4+1) Para resolverla, aplicamos un criterio similar al problema inicial, pero restando los 2 primeros sumandos y luego, todo *x. Notaremos que se forman Ln y arctan. Saludos.
Sir please do a limit question which was came in JEE Advanced 2014 shift-1 question number 57 it's a question of a limit lim as x-->1 [{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4 You have to find the greatest value of a It has 2 possible answers 0 and 2 But I want the reason that why should I reject 2 and accept 0 Because final answer is 0 Please help 😢
2 cannot be a solution. If you substitute a = 2, you get: lim [(-1)^(1+rt(x))] x->1 given that (-1)^n is only defined in R for n (E Z, we cannot compute this limit
Sir why not from the beginning we make the denominator like summation of two squares as x⁴+1 can be written (x²)²+1² and use arctan rule sorry I'm still in high school and haven't studied this law thoroughly. I just learned about it for the first time from your previous video so I don't know much about it sorry
We could've easily done this by multiplying and dividing by 2 and then separating it to 1+1 and adding x²-x² and then splitting it, taking x² common and then we'll have it!
@@made-bp6uf Thank you so much! I got tan-1 and tanh-1 functions with complex terms by factoring (1+ix^2) and (1-ix^2). And, I found the exponential expression of inverse functions of tan-1 and tanh-1. But the result cannot be derived to be equal to the solution of this video ;) plz help me!
Your sheer delight in algebraic manipulation is infectious. Thank-you.
Plus sign between two arctan, thanks. Very clear explanation.
You are great man. With a smile after each step, you make the whole solution exciting
Excellent, sir. Instead of K, you can use Y. K is normally used as a constant. 1/4SQRT2 x SQRT2= 1/4. S Chitrai Kani
The roots of the denominator are just the 4th roots of -1, which we can group into complex conjugate pairs (r1, r2) = e^(±pi*i/4) and (r3, r4) = e^(±3*pi*i/4).
(x-r1)*(x-r2) = x^2 + 2*cos(pi/4)*x + 1 = x^2 + √2*x + 1
(x-r3)*(x-r4) = x^2 + 2*cos(3*pi/4)*x + 1 = x^2 - √2*x + 1
therefore, x^4 + 1 = (x^2 + √2*x + 1)*(x^2 - √2*x + 1)
Also, if you don't mind having logarithms with complex arguments and coefficients in the final result, the problem becomes very straightforward.
The result being r/4 * (ln((x+r)/(x-r)) + i*ln((x+i*r)/(x-i*r))) where r = e^(pi*i/4)
So fun to watch you make maths enjoyable!
Wooow , dear professor you are so patient ❤❤
This solution is very long and I'd prefer first method
Divide both numerator and denominator by x^2. Then by algebraic manipulations, we can write the ratio as sum of two ratios which can be integrated individually.
Great ! Audible voice in facilitating and I like like the way how you manipulate algebra❤
22:33 - Where is the addition sign between the two terms tan^-1 ?
He might’ve forgot it. I was curious about it too.
@@baymax8984 yes bro . I thought the same thing but I thought it would be better to point it out and ask questions.
He forgot it
The process of manipulation of numerators can be made tidier by using polynomial division: divide numerator with the derivative of the denominator polynomial and then i t naturally splits.
The beginning is stunning. 🤩
Podemos reescribir:
1/(x^4+1)=[1/2*(x^2+1-√2 x)+1/2*(x^2+1+√2 x)]-x^2]/(x^4+1)
Luego, al integrar, separamos los 3 sumandos. Notaremos que las 2 primeras integrales son directas:
I1=int 1/2*1/[(x+√2 x)^2+(√2/2)^2] -> arctan
Igualmente, I2 -> arctan
I3=int x^2/(x^4+1)
Para resolverla, aplicamos un criterio similar al problema inicial, pero restando los 2 primeros sumandos y luego, todo *x. Notaremos que se forman Ln y arctan.
Saludos.
Sir please do a limit question which was came in
JEE Advanced 2014 shift-1 question number 57
it's a question of a limit
lim as x-->1
[{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4
You have to find the greatest value of a
It has 2 possible answers 0 and 2
But I want the reason that why should I reject 2 and accept 0
Because final answer is 0
Please help 😢
Okay. I'll check it out
It's really hard
2 cannot be a solution.
If you substitute a = 2, you get:
lim [(-1)^(1+rt(x))]
x->1
given that (-1)^n is only defined in R for n (E Z, we cannot compute this limit
Excellent licture're
Just the other day a student of mine asked this to me. Was a fun challenge
Amazing sir
Substitute x for sqrt of tabhebt, maybe?
x²-rad(2)x+1 is, indeed, always positive
Hi we would be glad if you
solve this using complex analysis aka the resideu theorem.
Of course if the integral is definated from 0 to infinity.
Ty.
This calc 3?
awesome vid. thanks!
Sir why not from the beginning we make the denominator like summation of two squares as x⁴+1 can be written (x²)²+1² and use arctan rule sorry I'm still in high school and haven't studied this law thoroughly. I just learned about it for the first time from your previous video so I don't know much about it sorry
Here should be + between arctans
These two arctans also can be combined
arctan(sqrt(2)x-1)+arctan(sqrt(2)x+1) = arctan(sqrt(2)x/(1-x^2))
Beat me to it…though he has the + in the answer written above it, before he simplifies the natural log terms. It appears to be a simple oversight.
Why its not tan inverse xsquare
I hope the third method is simple because I did not understand the solution to either of the two methods.😅😅😅😅😅
We could've easily done this by multiplying and dividing by 2 and then separating it to 1+1 and adding x²-x² and then splitting it, taking x² common and then we'll have it!
Would you pleaseeeeee be so kind to use -C in a video pleaseeee
Integration of x^n/1+x^4 is possible for all natural numbers of n,someone please correct me if i am rong
I'm trying to solve this problem by complex factoring... but it's not so easy...
Please help me..
what's your problem I'll be happy to help
@@made-bp6uf Thank you so much! I got tan-1 and tanh-1 functions with complex terms by factoring (1+ix^2) and (1-ix^2). And, I found the exponential expression of inverse functions of tan-1 and tanh-1. But the result cannot be derived to be equal to the solution of this video ;) plz help me!
Hi sir please help me integrate x^2*sqrt(x^2+1)
At 13:47 I don’t understand how 2sqrt(2) becomes sqrt(2)*sqrt(2).
i think that's root2 - root2
😮
∫ u/v •dx u=1; v=(1+x^4)
∫ u^2/2v^2/2
1^2/2[x+(1/5)x^5]^2/2-•dx
2/2[x+(1/5)x^5]^2 •dx
1/[(5x+x^5)/5]^2 •dx
5^2/(5x+x^5)^2 +c
∫ [5/5x+x^5]+c
Indians assemble here❤
Us bro us 😅
Thank you so much 🙏🏻❤
That
Was
Awesome