text: Q:
  
  Can anyone give me a hint on how to start this problem?
  
  Show that if $a > -1$ and $b > a+1$, then the integral
  $\int_0^{\infty} \frac{x^a}{1+x^b} \thinspace dx
   $
   is convergent.
  Not really sure how to start I tried to integrate but I'm not getting anywhere. 
  
  A:
  
  We have $1 + x^b \ge x^b$, so that:
  $$0 \le \int_1^{\infty} \frac{x^a}{1+x^b}dx \le \int_1^{\infty} \frac1{x^{b-a}}dx$$
  The rightmost integral is convergent because $b - a > 1$.
  On the other hand, note that $-a < 1$ and:
  $$0 \le \int_0^1 \frac{x^a}{1+x^b} dx \le \int_0^1 x^a dx = \int_0^1 \frac1{x^{-a}}dx$$
  
  