Integral:(1+x^(-2/3))^(1/2)dx, Solve By Substitution

[dagr8est]

Integral1+x^(-2/3))^(1/2)dx

Just started integration and I need to solve this by substitution but I don't see how I can with this problem. Am I just not seeing it or is there no way to solve by substitution? Any help would be appreciated.

 

[galactus]

Let \(\displaystyle \L\\u^{2}=1+x^{\frac{2}{3}}\)

\(\displaystyle \L\\x^{\frac{2}{3}}=u^{2}-1\)

\(\displaystyle \L\\x^{\frac{-2}{3}}=\frac{1}{u^{2}-1}\)

\(\displaystyle \L\\x=(\frac{1}{u^{2}-1})^{\frac{-3}{2}}\)

$\displaystyle dx=\frac{3u}{\sqrt{\frac{1}{u^{2}-1}}}$

Therefore, we have:

\(\displaystyle \L\\\int{\sqrt{1+\frac{1}{u^{2}-1}}}\frac{3u}{\sqrt{\frac{1}{u^{2}-1}}}\)

Skipping a little algebra, this boils down to

\(\displaystyle \L\\\int{3u^{2}}du\)

 

[dagr8est]

Thanks for the help.

 

[soroban]

Hello, dagr8est!

Here's another approach . . .

\(\displaystyle \L \int \left(1\,+\,x^{-\frac{2}{3}}\right)^{\frac{1}{2}}\,dx\)

Under the radical we have: $\displaystyle \,1\,+\,x^{-\frac{2}{3}} \;= \;1\,+\,\frac{1}{x^{\frac{2}{3}}} \;=\;\frac{x^{\frac{2}{3}}\,+\,1}{x^{\frac{2}{3}}}$

Then we have: \(\displaystyle \L\,\sqrt{\frac{x^{\frac{2}{3}}\,+\,1}{x^{\frac{2}{3}}}} \;= \;\frac{\sqrt{x^{\frac{2}{3}}\,+\,1}}{x^{\frac{1}{3}}}\)

The integral becomes: \(\displaystyle \L\,\int\)$\displaystyle x^{-\frac{1}{3}}\left(x^{\frac{2}{3}}\,+\,1\right)^{1/2}\,dx$

$\displaystyle \text{Now let }u\,=\,x^{\frac{2}{3}}\,+\,1$