Integration by substitution

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TODO: Flesh this page out with explanation and Jacobian reference



1 dimensional example

Question: 304x(1+x2)2dx

That 1+x2 is the problem, and the derivative is clearly something to do with just x (infact =2x) so this invites a substitution.

  • Substitute u=1+x2 then dudx=2x thus du=2xdx
    • Notice our integral is I=304x(1+x2)2dx=302(1+x2)22xdx we can literally replace the 2xdx with du - this is a nice shortcut.
    • More formally we actually have dx=du2x let us substitute this
      304x(1+x2)2dx=x=3x=04xu2du2x
      Notice: I write explicitly the limits of the integral because I have yet to express them in terms of u
      =x=3x=02u2du=2x=3x=01u2du
      • Converting the limits:
        • x=3u=1+32=10
        • x=0u=1
      • Thus I=21011u2du which the reader ought to be able to integrate.