Convergence of Integral of 1/sqrt(x+x^3) using Limit Comparison Test

Question

I tried to get the answer by using the LCT, and it led me to the cpnclusion that the integral diverges since the integral from 0 to 1 diverges, then the whole integral also does it

Answer 1

The limit comparison test usually is used for the convergence (or lack thereof) of series

 

The comparison test can be used to show that the improper integral

 

 

∫₀^∞  ( x + x³)^-1/2 dx  converges

 

 

One way to do this is to split the range of integral into [0,1]  and (1,∞)   and show that both 

 

   I1  =   ∫₀¹ ( x + x3)^-1/2 dx converges  and that

 

I2  =  ∫₁^∞ ( x + x3)^-1/2 dx converges

 

The integral I1  is less than  ∫₀¹  dx / sqrt(x) which is easily evaluated to be 2

 

The inegral I2  is less than  ∫₁^∞  dx x^-3/2   which is easily evaluated to be also equal to 2.  

Thus the original integral will be less than 4  Since it cannot be negative, this proves convergence.