Question

Question #6d568

Answer 1

I do not know why it says to consider separate cases; the integration that I did (below) seems to work in both cases.

Explanation:

Given: `int(1+x^(-2/3))^(1/2)dx`

`int(1+1/x^(2/3))^(1/2)dx=`

`int((x^(2/3)+1)/x^(2/3))^(1/2)dx=`

`int((x^(2/3)+1))^(1/2)/x^(1/3)dx=`

`intsqrt(x^(2/3)+1)/x^(1/3)dx`

Let `u = x^(2/3)", then "du = 2/3u^(-1/3)dx" or "3/2du= u^(-1/3)dx`

`3/2intsqrt(u+1)du = (u+1)^(3/2) + C`

Reverse the substitution:

`int(1+x^(-2/3))^(1/2)dx = (x^(2/3)+1)^(3/2) + C`