## Improper Integral

1. The problem statement, all variables and given/known data
Evaluate the integral: $$\int$$$$\frac{dx}{x^{3}+x^{2}+x+1}$$
from infinity to zero

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

3. 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!
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 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?
 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

## Improper Integral

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.
 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

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