evaluate the integral or state that it diverges: integral from 0 to 1 of ((x+1)/sqrt((x^2)+2x))dx
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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.
the problem has been misread, the integrand is actually $\displaystyle \frac{x+1}{\sqrt{x^2+2x}}.$
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 $\displaystyle \int_{0}^1\frac{(x+1)}{\sqrt{x^2+2x}}dx$ let $\displaystyle u=x^2+2x$ so $\displaystyle du=2(x+1)dx$ $\displaystyle \int_{0}^3\frac{(x+1)}{2(x+1)\sqrt{u}}du$ $\displaystyle \frac{1}{2}\int_{0}^3u^{-\frac{1}{2}}du$
just to mention that the integral converges by limit comparison test with $\displaystyle \int_0^1\frac{dx}{\sqrt x}.$
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