Question #6d568

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Answer 1 · 2017-04-17T06:49:46

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

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#