The integral of $\displaystyle x^3/\sqrt(1+x)$

How do you start this integral since it is an improper fraction. I tried squaring the top and bottom, but that didn't seem to work. What else should I try

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- Dec 5th 2009, 05:59 PMstevo970integral with improper fraction
The integral of $\displaystyle x^3/\sqrt(1+x)$

How do you start this integral since it is an improper fraction. I tried squaring the top and bottom, but that didn't seem to work. What else should I try - Dec 5th 2009, 06:01 PMJhevon
- Dec 5th 2009, 06:01 PMstevo970
correction $\displaystyle x^3/\sqrt(1+x^2)$

- Dec 5th 2009, 06:03 PMJhevon
- Dec 6th 2009, 02:07 PMstevo970
so would you then have to figure out

du=$\displaystyle x/\sqrt(1+x^2)$dx or is that not the next step - Dec 6th 2009, 02:24 PMJhevon