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amyna

  • one year ago

find the intergal: Don't know what "u" is in this problem?

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  1. amyna
    • one year ago
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    |dw:1441849092988:dw|

  2. jim_thompson5910
    • one year ago
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    let `u = arctan(x)` then `du/dx = 1/(1+x^2)` ----> `du = dx/(1+x^2)`

  3. amyna
    • one year ago
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    |dw:1441849417744:dw| i got this, then i don't know what to do?

  4. zepdrix
    • one year ago
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    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.

  5. zepdrix
    • one year ago
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    You decided that,\[\large\rm \color{royalblue}{u=\arctan x},\qquad\qquad du=?\]

  6. amyna
    • one year ago
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    du= 1/1+x^2

  7. zepdrix
    • one year ago
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    Good. Don't forget about your differential though! :)

  8. zepdrix
    • one year ago
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    \[\large\rm \color{royalblue}{u=\arctan x},\qquad\qquad \color{orangered}{du=\frac{1}{1+x^2}dx}\]

  9. zepdrix
    • one year ago
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    \[\large\rm \int\limits\limits \color{royalblue}{\arctan x}\left(\color{orangered}{\frac{1}{1+x^2}dx}\right)=\int\limits\limits \color{royalblue}{u}\left(\color{orangered}{du}\right)\]Did the colors help? :o Not so much?

  10. amyna
    • one year ago
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    yes! Thanks!!

  11. amyna
    • one year ago
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    is the answer arctan^2(x)/2 }+c?

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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