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Thread: 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

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

    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

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

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

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

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


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

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

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

    So the integral becomes

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

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

    $\displaystyle = \int{1\,d\theta}$

    $\displaystyle = \theta + C$

    $\displaystyle = \arctan{x} + C$.


    So $\displaystyle \int{1 - \frac{1}{1 + x^2}\,dx} = x - \arctan{x} + C$.
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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.

    $\displaystyle \frac{x^2}{1 + x^2} = \frac{1 + x^2 - 1}{1 + x^2} = \frac{1 + x^2}{1 + x^2} - \frac{1}{1 + x^2} = 1 - \frac{1}{1 + x^2}$.
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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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    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 $\displaystyle u = x^2 $ so $\displaystyle du=2xdx$ 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 $\displaystyle \arctan{x} \neq \ln{(1 + x^2)}$.

    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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    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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    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:

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


    We can also see that

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

    This comes from a $\displaystyle u$ substitution.

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


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

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

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

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

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



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

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

    Watch what happens.

    Let $\displaystyle u = 1 + x^2$.

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


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

    So the rule for $\displaystyle \ln$ 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 $\displaystyle \arctan{x}$ and seeing if we get $\displaystyle \frac{1}{1 + x^2}$. If we do, then $\displaystyle \int{\frac{1}{1 + x^2}\,dx} = \arctan{x}$.


    $\displaystyle y = \arctan{x}$

    $\displaystyle \tan{y} = x$

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

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

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

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


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


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

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

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

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


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

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


    No more arguments please!
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  15. #15
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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

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

    $\displaystyle u = (1+x^2)$

    $\displaystyle du = 2xdx$

    so

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