Improper Integrals 2

• May 20th 2010, 06:16 PM
smartartbug
Improper Integrals 2
evaluate the integral or state that it diverges:

integral from 0 to 1 of ((x+1)/sqrt((x^2)+2x))dx
• May 20th 2010, 06:34 PM
lilaziz1
Partial fractions should do the job:

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

$A(x+2) + Bx = x+1$

set x = -2 to get B; set x = 0 to get A.
• May 20th 2010, 08:04 PM
Krizalid
the problem has been misread, the integrand is actually $\frac{x+1}{\sqrt{x^2+2x}}.$
• May 20th 2010, 08:28 PM
DrDank
Quote:

Originally Posted by smartartbug
evaluate the integral or state that it diverges:

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

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

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

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

$\frac{1}{2}\int_{0}^3u^{-\frac{1}{2}}du$
• May 20th 2010, 08:59 PM
Krizalid
just to mention that the integral converges by limit comparison test with $\int_0^1\frac{dx}{\sqrt x}.$