# Improper Integrals 2

#### smartartbug

evaluate the integral or state that it diverges:

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

#### lilaziz1

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.

smartartbug

#### Krizalid

MHF Hall of Honor
the problem has been misread, the integrand is actually $$\displaystyle \frac{x+1}{\sqrt{x^2+2x}}.$$

AllanCuz

#### DrDank

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 $$\displaystyle du=2(x+1)dx$$

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

$$\displaystyle \frac{1}{2}\int_{0}^3u^{-\frac{1}{2}}du$$

HallsofIvy

#### Krizalid

MHF Hall of Honor
just to mention that the integral converges by limit comparison test with $$\displaystyle \int_0^1\frac{dx}{\sqrt x}.$$

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