Integral of 1/(x^2+2x+1) (substitution)
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- Опубликовано: 9 фев 2025
- Integral of 1/(x^2+2x+1) (substitution) - How to integrate it step by step using the substitution method!
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Derivative of -1/(x+1) = • Derivative of -1/(x+1)
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WWonderful video.
Thank you very much! ❤❤
Nice.
Thanks! ❤
Good.
Thanks! ❤❤
Hi, can you please explain how did you know that it is impossible to do it with fraction decomposition without trying it first ?
HI! Because you cannot write x^2+2x+1 as (x-a)(x-b) with a different from b. If the expression in the denominator can be written as (x+a)^2 or a+(x+b)^2 (a>0), then we cannot solve it by partial fraction. Said in another way, we can only use partial fraction decomposition when we get two different solutions from the second degree formula
Sir plzzzzzzzzzzzzzzzz I need your help !
Here integration of 1/(2x+1)(x+2)
Using partial fraction (without substitution) I get value of A = 2/3 , B =1/3
(By solving As we get 2 equation here
A +2B=0 -----> 1 equation
2A+ B=1 ---> 2 equation
When we put B= -1/3 in 2nd equation we get A = 1/3 which is correct but when we put B =-1/3 in 1 equation we get A =2/3 which is incorrect , it should be same in both case , so why it is not same here ?)
And
When using partial fraction (using substitution )i get value A =1/3 and B = -1/3 which is our correct answer so ,
My doubt is why I am getting A = 2/3 when done without substitution ? Why I am not getting 1/3 ? And how I will come to know that in which question substitution is required in partial fraction and in which question it can be done without substitution .
Hi, Jee Main! First of all, with your 2 equations I always get A=2/3 and B=-1/3 :
A+2B=0
2A+B=1
If A=2/3:
2/3 + 2B = 0 ==> 2B = -2/3 ==> B = (-2/3)/2 = -1/3
2(2/3) + B = 1 ==> 4/3 + B = 1 ==> B = 1 - 4/3 ==> B = 3/3 - 4/3 ==> B = -1/3
If B=-1/3:
A + 2(-1/3) = 0 ==> A - 2/3 = 0 ==> A = 2/3
2A + (-1/3) = 1 ==> 2A - 1/3 = 1 ==> 2A = 1 + 1/3 ==> 2A = 3/3 + 1/3 ==> 2A = 4/3 ==> A = (4/3)/2 ==> A = 2/3
For the second case, I don't know exactly the substitution method you are using, but I imagine that the begining you did something like 2x+1 = 2(x+1/2) and you did partial fraction decomposition without considering the 2 multiplying (x+1/2):
1/(2x+1)(x+2) = A/(2x+1) + B/(x+2) ==> A=2/3 and B=-1/3
(2/3)/(2x+1) + (-1/3)/(x+2) =
= (2/3)(x+2)/(2x+1)(x+2) + (-1/3)(2x+1)/(2x+1)(x+2) =
= [ (2/3)x + 4/3 ]/(2x+1)(x+2) + [ (-2/3)x - 1/3 ]/(2x+1)(x+2) =
= [ (2/3)x + 4/3 - (2/3)x - 1/3 ]/(2x+1)(x+2) =
= [ 3/3 ]/(2x+1)(x+2) =
= 1/(2x+1)(x+2)
1/2(x+1/2)(x+2) = A/(x+1/2) + B/(x+2) ==> A=1/3 and B=-1/3
(1/3)/(x+1/2) + (-1/3)/(x+2) =
= (2/3)/2(x+1/2) + (-1/3)/(x+2) =
= (2/3)/(2x+1) + (-1/3)/(x+2) =
= from here I did exactly the same as before =
= (2/3)(x+2)/(2x+1)(x+2) + (-1/3)(2x+1)/(2x+1)(x+2) =
= [ (2/3)x + 4/3 ]/(2x+1)(x+2) + [ (-2/3)x - 1/3 ]/(2x+1)(x+2) =
= [ (2/3)x + 4/3 - (2/3)x - 1/3 ]/(2x+1)(x+2) =
= [ 3/3 ]/(2x+1)(x+2) =
= 1/(2x+1)(x+2)
I think the problem is similar to factorize the polynomial P(x)=2x+1:
If you do 2x+1=0 you obtain x=-1/2 but when we write the polynomial we add the "2" multiplying:
P(x)=2(x-(-1/2)) = 2(x+1/2)
Same when factorizing P(x) = ax^2 + bx + c when "a" different from 1:
If we have P(x) = x^2 - 3x + 2 then, using the famous formula, we get x=1 and x=2.
Then P(x) = (x-1)(x-2)
But if we have Q(x) = 2x^2 - 6x + 4 (I've just multiplied by 2 P(x)), using the same formula we will also get x=1 and x=2.
However, we have to multiply by "a", in this case a=2:
Q(x) = 2(x-1)(x-2)
Hope it is clear now, and if not, ask me again with more details about the substitution you did ;)
@@IntegralsForYou
Here is the answear how I did using substitution
s.yimg.com/tr/i/c500bfc7d9194195a290dc5488dafacd_A.jpeg
My doubt is why answer is not coming when subsitution is not used, is it compulsary to use substitution here ?If yes why ?
Without substitution , we get A = 2/3 , B = -1/3
So from starting my doubt was why we are not getting A = 1/3 when substitution is not used which is the right answer given in my book
In both cases you have the right answer. If we do substitution, we get another function depending on "t". Then we will get a different solution, but if we undo the substitution we should get the same solution or equivalent (same derivative):
1. With substitution:
Integral of 1/[ (2x+1)(x+2) ] dx =
Substitution:
t = 2x + 1 ==> t-1 = 2x ==> x=(t-1)/2
dt = 2 dx ==> dt/2 = dx
= Integral of 1/[ t( (t-1)/2 + 2) ] dt/2 =
= Integral of 1/[ t( t - 1 + 4) ] dt =
= Integral of 1/[ t( t + 3) ] dt =
Partial fraction decomposition for 1/t(t+3):
1/t(t+3) = A/t + B/(t+3) =
= A(t+3)/t(t+3) + Bt/(t+3)t =
= (At + 3A + Bt)/t(t+3) =
= ( (A+B)t + 3A )/t(t+3)
==>
0 = A+B ==> B = -A = 1/3
1 = 3A ==> A = 1/3
==>
A = 1/3
B = -1/3
==>
1/t(t+3) = A/t + B/(t+3) = (1/3)/t - (1/3)/(t+3)
= Integral of 1/[ t( t + 3) ] dt =
= Integral of [ (1/3)/t - (1/3)/(t+3) ] dt =
= Integral of (1/3)/t dt - Integral of (1/3)/(t+3) dt =
= (1/3)Integral of 1/t dt - (1/3)Integral of 1/(t+3) dt =
= (1/3)ln|t| - (1/3)ln|t+3| =
= (1/3)ln|2x+1| - (1/3)ln|2x+1+3| =
= (1/3)ln|2x+1| - (1/3)ln|2x+4| + C
2. Without substitution:
Integral of 1/[ (2x+1)(x+2) ] dx =
Partial fraction decomposition for 1/(2x+1)(x+2):
1/(2x+1)(x+2) = A/(2x+1) + B/(x+2) =
= A(x+2)/(2x+1)(x+2) + B(2x+1)/(2x+1)(x+2) =
= (Ax + 2A + 2Bx + B)/(2x+1)(x+2) =
= ( (A+2B)x + (2A+B) )/(2x+1)(x+2)
==>
0 = A+2B ==> 0 = A + 2(1-2A) ==> 0 = A + 2 - 4A ==> 0 = -3A + 2 ==> 3A = 2 ==> A = 2/3
1 = 2A+B ==> B = 1-2A
==>
A = 2/3
B = 1 - 4/3
==>
A = 2/3
B = -1/3
==>
1/(2x+1)(x+2) = A/(2x+1) + B/(x+2) = (2/3)/(2x+1) - (1/3)/(x+2)
= Integral of [ 1/(2x+1)(x+2) ] dt =
= Integral of [ (2/3)/(2x+1) - (1/3)/(x+2) ] dt =
= Integral of (2/3)/(2x+1) dx - Integral of (1/3)/(x+2) dx =
= (1/3)Integral of 2/(2x+1) dx - (1/3)Integral of 1/(x+2) dx =
= (1/3)ln|2x+1| - (1/3)ln|x+2| + C
We have these solutions:
With substitution: (1/3)ln|2x+1| - (1/3)ln|2x+4| + C
Without substitution: (1/3)ln|2x+1| - (1/3)ln|x+2| + C
There are two questions here:
1. Are they correct solutions for the integral of 1/(2x+1)(x+2)?
2. Are they equal or equivalent?
1. Let's see the first one: check that the derivative is 1/(2x+1)(x+2):
Derivative of (1/3)ln|2x+1| - (1/3)ln|2x+4| + C =
= (1/3)(1/(2x+1))*2 - (1/3)(1/(2x+4))*2 + 0 =
= (2/3)((2x+4)/(2x+1)(2x+4)) - (2/3)((2x+1)/(2x+4)(2x+1)) =
= [ (4/3)x + 8/3 ]/(2x+1)(2x+4) - [ (4/3)x + 2/3 ]/(2x+4)(2x+1) =
= [ (4/3)x + 8/3 - (4/3)x - 2/3 ]/(2x+4)(2x+1) =
= 2/(2x+4)(2x+1) =
= 2/2(x+2)(2x+1) =
= 1/(x+2)(2x+1) ==> OK
Derivative of (1/3)ln|2x+1| - (1/3)ln|x+2| + C =
= (1/3)(1/(2x+1))*2 - (1/3)(1/(x+2)) + 0 =
= (2/3)((x+2)/(2x+1)(x+2)) - (1/3)((2x+1)/(x+2)(2x+1)) =
= [ (2/3)x + 4/3 ]/(2x+1)(x+2) - [ (2/3)x + 1/3 ]/(x+2)(2x+1) =
= [ (2/3)x + 4/3 - (2/3)x - 1/3 ]/(x+2)(2x+1) =
= 1/(x+2)(2x+1) ==> OK
2. They are equivalent. We will use the formula ln(ab) = ln(a) + ln(b)
Solution with substitution =
= (1/3)ln|2x+1| - (1/3)ln|2x+4| + C =
= (1/3)ln|2x+1| - (1/3)ln|2(x+2)| + C =
= (1/3)ln|2x+1| - (1/3)[ ln|2| + ln|x+2| ] + C =
= (1/3)ln|2x+1| - (1/3)ln|2| - (1/3)ln|x+2| + C =
= (1/3)ln|2x+1| - (1/3)ln|x+2| + C - (1/3)ln|2| =
= (1/3)ln|2x+1| - (1/3)ln|x+2| + C' = ( where C' = C - (1/3)ln|2| )
= Solution without substitution
We say that two solutions of an integral are equivalent when they are only different by the constant. Since the derivative of a constant is zero, both solutions have the same derivative.
Conclusion: you can choose doing or not the substitution, the only difference is that you will have an equivalent solution. Personally, in this case, I prefer not doing substitution, it only adds more steps without doing the resolution easier.
;-D
@@IntegralsForYou thank you very much sir 👍👍👍👍👍
Best teacher on RUclips
And subscribed 😊.
;-D I think I wrote you the longest answer in a comment I have :D. It was a really good question! Thanks for your comment ;-D
Hi, can you please do the integral of x/(x^2 -x +1)
Hi Anthony Khoury, here it is:
Integral of x/(x^2-x+1) dx =
= (1/2)Integral of 2x/(x^2-x+1) dx =
= (1/2)Integral of (2x-1+1)/(x^2-x+1) dx =
= (1/2)Integral of (2x-1)/(x^2-x+1) dx + (1/2)Integral of 1/(x^2-x+1) dx =
= (1/2)ln|x^2-x+1| + (1/2) ruclips.net/video/8I3g1rwrkpk/видео.html =
= (1/2)ln|x^2-x+1| + (1/2)(2/sqrt(3))arctan((2x-1)/sqrt(3)) =
= (1/2)ln|x^2-x+1| + (1/sqrt(3))arctan((2x-1)/sqrt(3)) + C
;-D
Hii solve it 'Integration x+1/1-2x-x^2
Hi Ankit Kushwaha! Here it is:
Integral of (x+1)/(1-2x-x^2) dx =
= Integral of (x+1)/(-x^2-2x+1) dx =
= Integral of (x+1)/-(x^2+2x-1) dx =
= -Integral of (x+1)/(x^2+2x-1) dx =
= -(1/2)Integral of 2(x+1)/(x^2+2x-1) dx =
= -(1/2)Integral of (2x+2)/(x^2+2x-1) dx =
= -(1/2)Integral of 1/(x^2+2x-1) (2x+2)dx =
Substitution:
u = x^2+2x-1
du = (2x+2)dx
= -(1/2)Integral of 1/u du =
= -(1/2)ln|u| =
= -(1/2)ln|x^2+2x-1| + C
;-D
Ur amazing thank u