Improper Integrals 2

Dec 2009
11
0
evaluate the integral or state that it diverges:

integral from 0 to 1 of ((x+1)/sqrt((x^2)+2x))dx
 
Mar 2010
107
14
Partial fractions should do the job:

\(\displaystyle \int_0^1 \frac{x+1}{x^2+2x}dx = \int_0^1 \frac{x+1}{x(x+2)}dx = \int_0^1 \left[\frac{A}{x} + \frac{B}{x+2}\right]dx \)

\(\displaystyle A(x+2) + Bx = x+1 \)

set x = -2 to get B; set x = 0 to get A.
 
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Krizalid

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Mar 2007
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the problem has been misread, the integrand is actually \(\displaystyle \frac{x+1}{\sqrt{x^2+2x}}.\)
 
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May 2010
20
8
evaluate the integral or state that it diverges:

integral from 0 to 1 of ((x+1)/sqrt((x^2)+2x))dx

\(\displaystyle \int_{0}^1\frac{(x+1)}{\sqrt{x^2+2x}}dx\)

let \(\displaystyle u=x^2+2x\) so \(\displaystyle du=2(x+1)dx\)

\(\displaystyle \int_{0}^3\frac{(x+1)}{2(x+1)\sqrt{u}}du\)

\(\displaystyle \frac{1}{2}\int_{0}^3u^{-\frac{1}{2}}du\)
 
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Krizalid

MHF Hall of Honor
Mar 2007
3,656
1,699
Santiago, Chile
just to mention that the integral converges by limit comparison test with \(\displaystyle \int_0^1\frac{dx}{\sqrt x}.\)
 
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