Solving Improper Integral: \int\frac{dx}{x^{3}+x^{2}+x+1} from 0 to ∞

[jen333]

Homework Statement

Evaluate the integral: $$\int$$$$\frac{dx}{x^{3}+x^{2}+x+1}$$
from infinity to zero

Homework Equations

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

The Attempt at a Solution

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


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!

 

[Pere Callahan]

Hi,
partial fractions, yes.

$$\frac{1}{x^3+x^2+x+1}=\frac{A}{x+1}+\frac{B+Cx}{x^2+1}=\frac{A(x^2+1)+(B+Cx)(x+1)}{x^3+x^2+x+1}$$

So you must have
$$1=A(x^2+1)+(B+Cx)(x+1)=(A+C)x^2+(B+C)x+A+B$$

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

Do you know to integrate the partial fractions?

 

[jen333]

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

 

[jen333]

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:
$$\frac{1/2}{x+1}$$-($$\frac{(1/2)x-(1/2)}{x^{2}+1}$$)

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

 

[Pere Callahan]

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