amyna
  • amyna
find the intergal: Don't know what "u" is in this problem?
Mathematics
katieb
  • katieb
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amyna
  • amyna
|dw:1441849092988:dw|
jim_thompson5910
  • jim_thompson5910
let `u = arctan(x)` then `du/dx = 1/(1+x^2)` ----> `du = dx/(1+x^2)`
amyna
  • amyna
|dw:1441849417744:dw| i got this, then i don't know what to do?

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zepdrix
  • zepdrix
I like to break it up like this :)\[\large\rm \int\limits \frac{\arctan x}{1+x^2}dx=\int\limits \color{royalblue}{\arctan x}\left(\color{orangered}{\frac{1}{1+x^2}dx}\right)\]Maybe the colors will help you to see what is going on.
zepdrix
  • zepdrix
You decided that,\[\large\rm \color{royalblue}{u=\arctan x},\qquad\qquad du=?\]
amyna
  • amyna
du= 1/1+x^2
zepdrix
  • zepdrix
Good. Don't forget about your differential though! :)
zepdrix
  • zepdrix
\[\large\rm \color{royalblue}{u=\arctan x},\qquad\qquad \color{orangered}{du=\frac{1}{1+x^2}dx}\]
zepdrix
  • zepdrix
\[\large\rm \int\limits\limits \color{royalblue}{\arctan x}\left(\color{orangered}{\frac{1}{1+x^2}dx}\right)=\int\limits\limits \color{royalblue}{u}\left(\color{orangered}{du}\right)\]Did the colors help? :o Not so much?
amyna
  • amyna
yes! Thanks!!
amyna
  • amyna
is the answer arctan^2(x)/2 }+c?

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