# Thread: comparison theroem

1. ## comparison theroem

Hi the instructions are to use the comparison thereom to determine whether the following integral is convergent or divergent

integral sign (S) sqrt x/(x^3 +1) dx

I am running into a mental road block and can't figure out how to get rid of the sq rt x as the numerator

What do I need to do to simplify this integral?

Thanks

Struggling calculus beginner

2. i see no bounds, but well.

i'll presume that is $\displaystyle 0\le x<\infty$ or $\displaystyle 1\le x<\infty.$

suppose is the first case, the split the integral into two pieces, one goes for $\displaystyle [0,1]$ and let's analyze the second piece for $\displaystyle x\ge1,$ then $\displaystyle \frac{\sqrt{x}}{x^{3}+1}<\frac{\sqrt{x}}{x^{3}}=\f rac{1}{x^{5/4}},$ hence the integral converges.

3. Sorry about my oversight

it is S a= 1 and b= infinity

so I assume that what you did is complete and I don't have to worry about the 0 to 1 section? right

Thanks

Calculus beginner

4. sure, since the function is integrable there 'cause is continuous on that compact set.