Help I can not solve this integral

  • Thread starter totentanz
  • Start date
  • #1
42
0
Can anyone help me solving x2+1/x4+1.........thanks
 

Answers and Replies

  • #2
22,089
3,297
Just apply linearity and you'll get:

[tex]\int{x^2dx}+\int{\frac{1}{x^4}dx}+\int{dx}[/tex].

This are elementary integrals, so it is easily solvable.
 
  • #3
42
0
thank you...but can you explain it more?
 
  • #4
22,089
3,297
What don't you understand about it? You just apply the formula

[tex]\int{(f(x)+g(x))dx}=\int{f(x)dx}+\int{g(x)dx}[/tex]

Then you have brought the integral to three basic integrals, which you can easily solve with the formula

[tex]\int{x^ndx}=\frac{x^{n+1}}{n+1}+C[/tex] for [tex]n\neq -1[/tex].
 
  • #5
133
1
Did you use partial fractions to get that result? I think he may be referring to:

[tex]
\int{\frac{{x^2}+1}{{x^4}+1}dx}
[/tex]
 
Last edited:
  • #6
22,089
3,297
Did you use partial fractions to get that result? I think he may be referring to the integral of ((x^2 + 1)/(x^4 + 1))dx

Ah yes, that may make more sense...

To the OP: Parantheses in mathematics are EXTREMELY important!!!! Don't add too few of them.
 
  • #7
133
1
So, which question is it? (Before I spend a year doing partial fractions, completing the square, substitution, and god knows what else)
 
  • #8
42
0
Did you use partial fractions to get that result? I think he may be referring to:

[tex]
\int{\frac{{x^2}+1}{{x^4}+1}dx}
[/tex]

Yes this is what I mean...but when you decompose the fraction I get

(x^2+1)/((x^2+1)(x^2+1)+2)
 
Last edited:
  • #11
lurflurf
Homework Helper
2,440
138
-decompose into partial fractions (with completed squares)
-obvious
inverse tangent
 
  • #12
HallsofIvy
Science Advisor
Homework Helper
41,833
964
[itex]x^4+ 1= 0[/itex] has no real roots but we can write [itex]x^4= -1[/itex] so that [itex]x^2= \pm i[/itex] and then get [itex]x= \frac{\sqrt{2}}{2}(1+ i)[/itex], [itex]x= \frac{\sqrt{2}}{2}(1- i)[/itex], [itex]x= \frac{\sqrt{2}}{2}(-1+ i)[/itex], and [itex]x= \frac{\sqrt{2}}{2}(-1- i)[/itex] as the four roots. We can pair those by cojugates:
[tex]\left(x- \frac{\sqrt{2}}{2}(1+ i)\right)\left(x- \frac{\sqrt{2}}{2}(1- i)\right)[/tex][tex]= \left((x- \frac{\sqrt{2}}{2})- i\frac{\sqrt{2}}{2}\right)\left((x- \frac{\sqrt{2}}{2})+ i\frac{\sqrt{2}}{2}\right)[/tex][tex]= (x- \sqrt{2}{2})^2+ \frac{1}{2}= x^2- \sqrt{2}x+ 1[/tex]
and
[tex]\left(x- \frac{\sqrt{2}}{2}(-1+ i)\right)\left(x- \frac{\sqrt{2}}{2}(-1- i)\right)[/tex][tex]= x^2+ \sqrt{2}x+ 1[/tex]

That is, the denominator [itex]x^4+ 1[/itex] factors as
[tex](x^2+ \sqrt{2}x+ 1)(x^2- \sqrt{2}x+ 1)[/tex]
and you can use partial fractions with those.
 
Last edited by a moderator:
  • #13
42
0
[itex]x^4+ 1= 0[/itex] has no real roots but we can write [itex]x^4= -1[/itex] so that [itex]x^2= \pm i[/itex] and then get [itex]x= \sqrt{2}{2}(1+ i)[/itex], [itex]x= \sqrt{2}{2}(1- i)[/itex], [itex]x= \sqrt{2}{2}(-1+ i)[/itex], and [itex]x= \sqrt{2}{2}(-1- i)[/itex] as the four roots. We can pair those by cojugates:
[tex]\left(x- \frac{\sqrt{2}}{2}(1+ i)\right)\left(x- \frac{\sqrt{2}}{2}(1- i)\right)[/tex][tex]= \left((x- \frac{\sqrt{2}}{2})- i\frac{\sqrt{2}}{2}\right)\left((x- \frac{\sqrt{2}}{2})+ i\frac{\sqrt{2}}{2}\right)[/tex][tex]= (x- \sqrt{2}{2})^2+ \frac{1}{2}= x^2- \sqrt{2}x+ 1[/tex]
and
[tex]\left(x- \frac{\sqrt{2}}{2}(-1+ i)\right)\left(x- \frac{\sqrt{2}}{2}(-1- i)\right)[/tex][tex]= x^2+ \sqrt{2}x+ 1[/tex]

That is, the denominator [itex]x^4+ 1[/itex] factors as
[tex](x^2+ \sqrt{2}x+ 1)(x^2- \sqrt{2}x+ 1)[/tex]
and you can use partial fractions with those.

Thanks bro...this is what I was looking for...I hope it works
 
Last edited:
  • #14
42
0
sorry my friend but it is not working
 
  • #15
dextercioby
Science Advisor
Homework Helper
Insights Author
13,077
641
Back in my high school days I knew tricks for these. For an anti-derivative sought in a subset of R excluding 0, we can make the following tricks:

[tex] \int \frac{x^2 +1}{x^4 +1} \, dx = \int \frac{1+\frac{1}{x^2}}{x^2 + \frac{1}{x^2}} \, dx = \int \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2 + \left(\sqrt{2}\right)^2} = ... [/tex]
 
  • #16
42
0
Back in my high school days I knew tricks for these. For an anti-derivative sought in a subset of R excluding 0, we can make the following tricks:

[tex] \int \frac{x^2 +1}{x^4 +1} \, dx = \int \frac{1+\frac{1}{x^2}}{x^2 + \frac{1}{x^2}} \, dx = \int \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2 + \left(\sqrt{2}\right)^2} = ... [/tex]

Thanks my friend,this is great...because I am stuck in this exo,and I can not go on
 
  • #17
HallsofIvy
Science Advisor
Homework Helper
41,833
964
You have been given a number of different suggestions (dextercioby's making the problem almost trivial) but keep saying you cannot do it- and showing no work at all. Have you tried what dextercioby suggested? What did you do and where did you run into trouble.
 
  • #18
42
0
You have been given a number of different suggestions (dextercioby's making the problem almost trivial) but keep saying you cannot do it- and showing no work at all. Have you tried what dextercioby suggested? What did you do and where did you run into trouble.

I am working on solving it all week,and I can not find a complete answer except what Mr.dextercioby show...and I already solve this "normalization of the wavefunction",thank you for your care
 
  • #19
42
0
I am working on solving it all week,and I can not find a complete answer except what Mr.dextercioby show...and I already solve this "normalization of the wavefunction",thank you for your care

Ok...I solve it
 

Related Threads on Help I can not solve this integral

  • Last Post
Replies
5
Views
621
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
6
Views
1K
Replies
22
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
20
Views
2K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
2
Views
997
Top