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Math Help - Calculus - Integration

  1. #1
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    Calculus - Integration

    how would I go about integrating:

    [x^2 / (1+x^2)] dx
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    Quote Originally Posted by bhuang View Post
    how would I go about integrating:

    [x^2 / (1+x^2)] dx
    use long division and get

    \int \bigg(1 - \frac{1}{1+x^2}\bigg)dx

    Then you should remember the second part is arctan
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    Quote Originally Posted by 11rdc11 View Post
    use long division and get

    \int (1 - \frac{1}{1+x^2})dx

    Then you should remember the second part is arctan
    And if you didn't know that, you could use trigonometric substitution.

    \int{1 - \frac{1}{1 + x^2}\,dx} = \int{1\,dx} - \int{\frac{1}{1 + x^2}\,dx}

     = x - \int{\frac{1}{1 + x^2}\,dx}.


    To find \int{\frac{1}{1 + x^2}\,dx}

    Let x = \tan{\theta} so that dx = \sec^2{\theta}\,d\theta.

    Also note that \theta = \arctan{x}.

    So the integral becomes

    \int{\frac{1}{1 + \tan^2{\theta}}\,\sec^2{\theta}\,d\theta}

     = \int{\frac{1}{\sec^2{\theta}}\,\sec^2{\theta}\,d\t  heta}

     = \int{1\,d\theta}

     = \theta + C

     = \arctan{x} + C.


    So \int{1 - \frac{1}{1 + x^2}\,dx} = x - \arctan{x} + C.
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    can you do the long division part?
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    Quote Originally Posted by bhuang View Post
    can you do the long division part?
    Of course you can.

    \frac{x^2}{1 + x^2} = \frac{1 + x^2 - 1}{1 + x^2} = \frac{1 + x^2}{1 + x^2} - \frac{1}{1 + x^2} = 1 - \frac{1}{1 + x^2}.
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    i haven't learned arctan or trigonometry for integration. so can I not simply integrate 1/(1+x^2) as ln(1+x^2)?
    so my final answer would be
    x-ln(1+x^2)+c
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    Super Member 11rdc11's Avatar
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    Quote Originally Posted by bhuang View Post
    i haven't learned arctan or trigonometry for integration. so can I not simply integrate 1/(1+x^2) as ln(1+x^2)?
    so my final answer would be
    x-ln(1+x^2)+c
    No since  u = x^2 so du=2xdx which you do not have in the numerator.
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    Quote Originally Posted by bhuang View Post
    i haven't learned arctan or trigonometry for integration. so can I not simply integrate 1/(1+x^2) as ln(1+x^2)?
    so my final answer would be
    x-ln(1+x^2)+c
    It should also be clear that \arctan{x} \neq \ln{(1 + x^2)}.

    I showed you how to solve it using trigonometric substitution...
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    I don't want to use something I haven't learned, it will confuse me before my test. I might not even remember it if I come across a similar question. I never made u=x^2.
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  10. #10
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by bhuang View Post
    I don't want to use something I haven't learned, it will confuse me before my test. I might not even remember it if I come across a similar question. I never made u=x^2.
    There isn't another way to solve this but the way Prove It and I showed you. What did you use as your u then?
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  11. #11
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    I didn't have a u. I went from 1-1/(1+x^2) and I integrated.
    1 becomes x, when integrated. And 1/(1+x^2) I made (1+x^2)^-1 so I could integrate that into ln(1+x^2).
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    Quote Originally Posted by bhuang View Post
    I didn't have a u. I went from 1-1/(1+x^2) and I integrated.
    1 becomes x, when integrated. And 1/(1+x^2) I made (1+x^2)^-1 so I could integrate that into ln(1+x^2).
    You aren't allowed to do that in math.
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    Why? What's wrong with doing that?
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    Watch carefully:

    \int{\frac{1}{x}\,dx} = \ln{|x|} + C by definition.


    We can also see that

    \int{\frac{1}{ax + b}\,dx} = \frac{1}{a}\ln{|ax + b|}.

    This comes from a u substitution.

    Let u = ax + b so that \frac{du}{dx} = a.


    So \int{\frac{1}{ax + b}\,dx} = \frac{1}{a}\int{\frac{a}{ax + b}\,dx}

     = \frac{1}{a}\int{\frac{1}{u}\,\frac{du}{dx}\,dx}

     = \frac{1}{a}\int{\frac{1}{u}\,du}

     = \frac{1}{a}\ln{|u|} + C

     = \frac{1}{a}\ln{|ax + b|} + C.



    BUT you are trying to find \int{\frac{1}{1 + x^2}\,dx}.

    We DO NOT have a formula for that yet because it is not of the form \frac{1}{ax + b}. But you could TRY a u substitution.

    Watch what happens.

    Let u = 1 + x^2.

    We can see that \frac{du}{dx} = 2x.


    Can we change \int{\frac{1}{1 + x^2}\,dx} into a function that is of the form \int{\frac{2x}{1 + x^2}\,dx}? NO, we can not.

    So the rule for \ln DOES NOT WORK in this case.


    Now see my post regarding trigonometric substitution to show you the method that DOES work.


    Alternatively, let's try differentiating \arctan{x} and seeing if we get \frac{1}{1 + x^2}. If we do, then \int{\frac{1}{1 + x^2}\,dx} = \arctan{x}.


    y = \arctan{x}

    \tan{y} = x

    \frac{d}{dx}(\tan{y}) = \frac{d}{dx}(x)

    \frac{dy}{dx}\,\frac{d}{dy}(\tan{y}) = 1

    \sec^2{y}\,\frac{dy}{dx} = 1

    \frac{dy}{dx} = \cos^2{y}.


    But we know that y = \arctan{x} and we also know that \cos{\theta} = \frac{1}{\sqrt{1 + \tan^2{\theta}}}


    So \frac{dy}{dx} = \left(\frac{1}{\sqrt{1 + \tan^2{y}}}\right)^2

     = \frac{1}{1 + \tan^2{y}}

     = \frac{1}{1 + (\tan{\arctan{x}})^2}

     = \frac{1}{1 + x^2}.


    So \frac{d}{dx}(\arctan{x}) = \frac{1}{1 + x^2}.

    Therefore \int{\frac{1}{1 + x^2}\,dx} = \arctan{x} + C.


    No more arguments please!
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  15. #15
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by bhuang View Post
    Why? What's wrong with doing that?

    ok so lets say we try to attempt it your way

    \int(1+x^2)^{-1}dx

    u = (1+x^2)

    du = 2xdx

    so

    \int u^rdu = \frac{u^{r+1}}{r+1} +c
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