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Integration of Rational Functions by Partial Fractions

renyikouniao

Member
Jun 1, 2013
41
1) integral (upper bound:1, lower bound:0) (x^2+1)/(x^3+x^2+4x) dx
2) integral (upper bound:1, lower bound:0) (x^4+x^2+1)/(x^3+x^2+x-3) dx

Now I know how to use Partial Fractions,My question is:

1) For the first part ln(x) is not defined at 0

¼ʃ1/x dx + ¼ʃ(3x-1)/(x²+x+4) dx
= ¼ ln|x| + ¼ʃ(3x-1)/(x²+x+4) dx



2) ln(x-1) is not defined at 1 for this part
ʃ1/[2(x-1)] + (x+7) / [2(x²+2x+3)] dx
= ½ʃ1/(x-1) +½ ʃ(x+7)/(x²+2x+3) dx
= ½ ln |x-1| +½ ʃ(x+7)/(x²+2x+3) dx


So If I want to evaluate this definite integral, what I should do next?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle \int^1_0\frac{x^2+1}{x^3+x^2+4x} dx\)

The first question that you would probably ask is whether the integral converges because the function is not defined at \(\displaystyle 0\).
 

renyikouniao

Member
Jun 1, 2013
41
\(\displaystyle \int^1_0\frac{x^2+1}{x^3+x^2+4x} dx\)

The first question that you would probably ask is whether the integral converges because the function is not defined at \(\displaystyle 0\).
Since the integral(upper bound: 1, lower bound: 0) 1/x is divergent, so the definite integral can't be evaluated?

Same as the second part, integral (upper bound:1, lower bound: 0) 1/(x-1) is divergent also.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Since the integral(upper bound: 1, lower bound: 0) 1/x is divergent, so the definite integral can't be evaluated?

Same as the second part, integral (upper bound:1, lower bound: 0) 1/(x-1) is divergent also.
Basically , before you integrate a definite integral you check for convergence . Sometimes you need to simplify a little bit before you make sure it is convergent .

If the integral is a sum of a convergent and divergent integral then it is divergent .