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Integral:(1+x^(-2/3))^(1/2)dx, Solve By Substitution
- Thread starter dagr8est
- Start date
1+x^(-2/3))^(1/2)dxLet \(\displaystyle \L\\u^{2}=1+x^{\frac{2}{3}}\)
\(\displaystyle \L\\x^{\frac{2}{3}}=u^{2}-1\)
\(\displaystyle \L\\x^{\frac{-2}{3}}=\frac{1}{u^{2}-1}\)
\(\displaystyle \L\\x=(\frac{1}{u^{2}-1})^{\frac{-3}{2}}\)
\(\displaystyle dx=\frac{3u}{\sqrt{\frac{1}{u^{2}-1}}}\)
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
\(\displaystyle \L\\\int{3u^{2}}du\)
\(\displaystyle \L\\x^{\frac{2}{3}}=u^{2}-1\)
\(\displaystyle \L\\x^{\frac{-2}{3}}=\frac{1}{u^{2}-1}\)
\(\displaystyle \L\\x=(\frac{1}{u^{2}-1})^{\frac{-3}{2}}\)
\(\displaystyle dx=\frac{3u}{\sqrt{\frac{1}{u^{2}-1}}}\)
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
\(\displaystyle \L\\\int{3u^{2}}du\)
Hello, dagr8est!
Here's another approach . . .
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: \(\displaystyle \L\,\int\)\(\displaystyle x^{-\frac{1}{3}}\left(x^{\frac{2}{3}}\,+\,1\right)^{1/2}\,dx\)
\(\displaystyle \text{Now let }u\,=\,x^{\frac{2}{3}}\,+\,1\)
Here's another approach . . .
Under the radical we have: \(\displaystyle \,1\,+\,x^{-\frac{2}{3}} \;= \;1\,+\,\frac{1}{x^{\frac{2}{3}}} \;=\;\frac{x^{\frac{2}{3}}\,+\,1}{x^{\frac{2}{3}}}\)\(\displaystyle \L \int \left(1\,+\,x^{-\frac{2}{3}}\right)^{\frac{1}{2}}\,dx\)
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: \(\displaystyle \L\,\int\)\(\displaystyle x^{-\frac{1}{3}}\left(x^{\frac{2}{3}}\,+\,1\right)^{1/2}\,dx\)
\(\displaystyle \text{Now let }u\,=\,x^{\frac{2}{3}}\,+\,1\)