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 😢
Your sheer delight in algebraic manipulation is infectious. Thank-you.
Plus sign between two arctan, thanks. Very clear explanation.
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
The beginning is stunning. 🤩
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 😢
Okay. I'll check it out
It's really hard
Just the other day a student of mine asked this to me. Was a fun challenge
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))
You’re the best!
Amazing sir
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
Why its not tan inverse xsquare
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
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.
Indians assemble here❤
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 🙏🏻❤