Can i do this to prove convergence of improper integral

[Dell]

given the improper integral from 0 to 1 of

$$\int$$dx/$$\sqrt[3]{x}$$(e^e-e^-x)

i am asked if it comverges or diverges,

what i have learned is that if i can find :
-a similar function that is bigger than my function and that i know converges, then my function also converges
-a similar function that is smaller than my function and that i know dicerges, then my function also diverges
-a function g(x) such that
lim f(x)/g(x)=K, (K not 0, not infinity) then f(x) acts like g(x)
x->inf

my similar function g(x) is 1/3$$\sqrt[3]{x}$$ which is smaller than f(x), BUT my g(x) converges, so that doesn't prove anything according to the 1st 2 conditions,

generally can i use the 3rd condition even though my integral is from 0 to 1 and not to infinity??
in this case that too doesn't work. what can i do from here??

if not how would you solve this?

 

[tiny-tim]

given the improper integral from 0 to 1 of

$$\int$$dx/$$\sqrt[3]{x}$$(e^e-e^-x)

Hi Dell!

(i assume you mean /(e^x - e-^x))

Hint: the only difficulty is at x = 0 …

so what is the integrand approximately equal to near x = 0?

 

[Dell]

my mistake, the e^e is meant to read e^x,,, don't know if that makes much of a difference

really can't find anything that helps me 1/(2x^4/3) is very close near x=0, but 1/(2x^4/3) diverges and it is bigger so it doesn't really help me,... unless i can say
lim fx/gx = 1
x->inf

can i do that since my integral is not to infinity,?? also I am not sure they ar similar when x gets vey big