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Improper Integral

  • Thread starter jen333
  • Start date
59
0
1. Homework Statement
Evaluate the integral: [tex]\int[/tex][tex]\frac{dx}{x^{3}+x^{2}+x+1}[/tex]
from infinity to zero

2. Homework Equations
lim t--> infinity [/tex] [tex]\int[/tex] [tex]\frac{dx}{x^{3}+x^{2}+x+1}[/tex]


3. The Attempt at a Solution

lim t-->infinity [/tex] [tex]\int[/tex] [tex]\frac{dx}{(x+1)(x^{2}+1}[/tex]


I'm stuck on where to go from here. I tried partial fractions, but can't seem to get it. any hints would be a great help!
 

Answers and Replies

Hi,
partial fractions, yes.

[tex]
\frac{1}{x^3+x^2+x+1}=\frac{A}{x+1}+\frac{B+Cx}{x^2+1}=\frac{A(x^2+1)+(B+Cx)(x+1)}{x^3+x^2+x+1}
[/tex]

So you must have
[tex]
1=A(x^2+1)+(B+Cx)(x+1)=(A+C)x^2+(B+C)x+A+B
[/tex]

Comparing coefficients of the same powers of x you get the equation:


1=A+B
0=B+C
0=A+C

which you can easily solve, I assume :smile:

Do you know to integrate the partial fractions?
 
Last edited:
59
0
Oh! I see, i must have miswritten something when i was doing partial fractions. Thank you so much for the help!

I 'll give it a shot and see what comes up
 
59
0
alright, so I've worked on solving this problem up to:

a=1/2 b=1/2 c=-1/2

so my integral terms would be:
[tex]\frac{1/2}{x+1}[/tex]-([tex]\frac{(1/2)x-(1/2)}{x^{2}+1}[/tex])

taking the antiderivative:
i have, [tex]\frac{1}{2}[/tex]ln|x+1| for the first term
as for the second, i know one of the terms will be tan[tex]^{-1}[/tex]x because of the denominator, but i'm having troubles with the numerator since I can't use substitution for it.
 
Split the second term into two. For the one with the x in the numerator you can use the substitution


u=x^2
du=2xdx

The first term (with the constant numerator).. well..you know how to do it:smile:
 

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