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)
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
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!
Your sheer delight in algebraic manipulation is infectious. Thank-you.
Plus sign between two arctan, thanks. Very clear explanation.
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)
Wooow , dear professor you are so patient ❤❤
This solution is very long and I'd prefer first method
So fun to watch you make maths enjoyable!
The beginning is stunning. 🤩
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
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.
Amazing sir
Just the other day a student of mine asked this to me. Was a fun challenge
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.
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!
You’re the best!
I hope the third method is simple because I did not understand the solution to either of the two methods.😅😅😅😅😅
sir i think you can also do this one by forcing integration by parts
Would you pleaseeeeee be so kind to use -C in a video pleaseeee
Why its not tan inverse xsquare
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..
∫ 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
At 13:47 I don’t understand how 2sqrt(2) becomes sqrt(2)*sqrt(2).
i think that's root2 - root2
That
Was
Awesome
Thank you so much 🙏🏻❤
Indians assemble here❤
Integrate[1/(1+x^4),x]=(-Ln(x^2-Sqrt[2]+1)+Ln(x^2+Sqrt[2]x+1)-2ArcTan[1-Sqrt[2]x]+2ArcTan[Sqrt[2]x+1])/(Surd[2,4])+C It’s in my head.