# Question #6d568

Apr 17, 2017

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: $\int {\left(1 + {x}^{- \frac{2}{3}}\right)}^{\frac{1}{2}} \mathrm{dx}$

$\int {\left(1 + \frac{1}{x} ^ \left(\frac{2}{3}\right)\right)}^{\frac{1}{2}} \mathrm{dx} =$

$\int {\left(\frac{{x}^{\frac{2}{3}} + 1}{x} ^ \left(\frac{2}{3}\right)\right)}^{\frac{1}{2}} \mathrm{dx} =$

$\int {\left(\left({x}^{\frac{2}{3}} + 1\right)\right)}^{\frac{1}{2}} / {x}^{\frac{1}{3}} \mathrm{dx} =$

$\int \frac{\sqrt{{x}^{\frac{2}{3}} + 1}}{x} ^ \left(\frac{1}{3}\right) \mathrm{dx}$

Let $u = {x}^{\frac{2}{3}} \text{, then "du = 2/3u^(-1/3)dx" or } \frac{3}{2} \mathrm{du} = {u}^{- \frac{1}{3}} \mathrm{dx}$

$\frac{3}{2} \int \sqrt{u + 1} \mathrm{du} = {\left(u + 1\right)}^{\frac{3}{2}} + C$

Reverse the substitution:

$\int {\left(1 + {x}^{- \frac{2}{3}}\right)}^{\frac{1}{2}} \mathrm{dx} = {\left({x}^{\frac{2}{3}} + 1\right)}^{\frac{3}{2}} + C$