Do My Improper Integral Solutions Converge or Diverge?

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1. Determine whether the following diverges or converges. If converges, evaluate it.
a) integral from 1 to infinity of: 1/(2x^2 + x) dx
b) integral from 0 to 1 of: 1/(2x^2 + x) dx

I just want to check my answers. I got a) diverges and b) converges with value of ln|3/2|. Do these answers sound right? I would appreciate some feedback, thanks.
 
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If you switch your answers for a) and b), it's right.
 
You had a 50/50 chance and you blew it ;)

How did you got to your conclusions?

Basically you need to find an upper/lower bound (i.e. a simpler integral that you know how to evaluate) and show that the bound converges/diverges then the original integral also converges/diverges.
 
i don't see how the answers are opposite lol maybe someone can show me a) so i can compare to my work
 
mat331760298 said:
i don't see how the answers are opposite lol maybe someone can show me a) so i can compare to my work

You haven't shown your work yet. If you do that maybe we can figure out what's wrong.
 
haha looked it over and when i did integration i had ln(x) + ln(2x+1) instead of ln(x) - ln(2x+1). makes sense now
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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