Simple Integral

  • #1

Homework Statement



Hello. I have a simple integral here that has been stumping me for the last 30 minutes. It appears that my basic integration skills have gotten very rusty.

Homework Equations



[tex]\int{x^3}\sqrt{1+x^2}dx[/tex]

The Attempt at a Solution


I am pretty sure a simple substitution will do, but I have yet to find it. I have tried simplifying the expression various ways, integration by parts, and I have also tried a few substitutions. Any ideas?
 

Answers and Replies

  • #2
Dick
Science Advisor
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u=1+x^2 will do it.
 
  • #3
jgens
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The substitution x=tan(u) also works.
 
  • #4
Thank you Dick and Jgens.

[tex]\int{x^3}\sqrt{1+x^2}dx[/tex]

Let u=1+x^2[tex]\rightarrow[/tex]du=2xdx

[tex]\int{x^3}\sqrt{1+x^2}dx=\frac{1}{2}\int(u-1)\sqrt{u}du[/tex]

[tex]\frac{1}{2}\int(u-1)\sqrt{u}du=\frac{1}{2}(\int{u^{\frac{3}{2}}}du-\int{u^{\frac{1}{2}}du)[/tex]

[tex]\int{u^{\frac{3}{2}}}du=\frac{2}{5}u^{\frac{5}{2}}[/tex]

[tex]\int{u^{\frac{1}{2}}du=\frac{2}{3}u^{\frac{3}{2}}[/tex]

[tex]\frac{1}{2}\int(u-1)\sqrt{u}du=\frac{2}{10}u^{\frac{5}{2}}-\frac{2}{6}u^{\frac{3}{2}}[/tex]


It appears I made a mistake somewhere. I believe my result should be multiplied by 1/2, and I can't find where I left that out.
 
  • #5
Nevermind. I guess the second part of the problem is incorrect. Mathematica has arrived at the same conclusion as this also. Thanks for all the help.
 

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