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Math Help - Integrating fractions

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    Integrating fractions

    What is the method of integrating functions like 1/(1 + x^2)?
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    You realize that is the derivative of \arctan(x).
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    \int \frac{1}{1+x^2}~dx = \tan^{-1} x +C

    Commit this to memory.
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    here's some more u should know by heart

     \displaystyle \int x^n \; dx = \frac {x^{n+1}}{n+1} + C

     \displaystyle \int \sqrt[n]x \; dx = \frac {n}{n+1}x \sqrt[n]x + C

    \displaystyle \int \frac {dx}{1+x^2 } =\arctan {x} + C

    \displaystyle \int \frac {dx}{x^2-a^2}=\frac {1}{2a} \ln \big{|} \frac {x-a}{x+a}\big{|} + C

    \displaystyle \int \frac{dx}{\sqrt{x^2 \pm a^2}}=\ln{\big{|}x+\sqrt{x^2\pm a^2\big{|}}+C

    \displaystyle \int a^x \; dx = \frac {a^x}{\ln {a}} +C


    and so on we use like 30 maybe "basic" integrals, like table integrals which u can use without calculating them and with some basic operations, and substitutions (and a lot of practice) u can solve probably (for sure) any integral that u can get to do in your school
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    Yeah, I've never done substitutions like that before. Can someone elaborate a little bit more on what they mean?
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    Okay, so if one is integrating with that function, from -2 to 2, the answer would be 126.8? I think I did it incorrectly...
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    the general trick is to comple the square and then perform a trig. substitution.
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    Quote Originally Posted by bobsanchez View Post
    Okay, so if one is integrating with that function, from -2 to 2, the answer would be 126.8? I think I did it incorrectly...
     \displaystyle \int _{-2} ^2 \frac {1}{1+x^2} \; dx =\tan^{-1 } {(x)}\mid_{-2} ^2 = \tan^{-1} {(2)} + \tan ^{-1} {(2)} =  2 \tan^{-1} {(2)} =2.2143
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    Yeah, I think I was off a little. XD
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    Also, with functions like (2-x)^6, what would be the best strategy? I mean, I know I could multiply it all out, but surely there's another method...?
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    Are you familiar with the Binomial Theorem?
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    Only vaguely.
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    Quote Originally Posted by bobsanchez View Post
    Also, with functions like (2-x)^6, what would be the best strategy? I mean, I know I could multiply it all out, but surely there's another method...?
    Let \displaystyle u = 2-x then [LaTeX ERROR: Convert failed]

    After some practice, you will just see by inspection:

    \displaystyle \int(ax+b)^n\;{dx} = \frac{1}{a(n+1)}(ax+b)^{n+1}+k

    Since by letting u = ax+b, we have \displaystyle \dfrac{du}{dx} = a \Rightarrow {dx} = \dfrac{du}{a} thus [LaTeX ERROR: Convert failed]

    (If a = -1, and b = 2, we have the above problem).
    Last edited by TheCoffeeMachine; August 9th 2010 at 11:02 PM.
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    Senior Member yeKciM's Avatar
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    Quote Originally Posted by bobsanchez View Post
    Also, with functions like (2-x)^6, what would be the best strategy? I mean, I know I could multiply it all out, but surely there's another method...?
    if u substitute  u = 2-x and  du=-dx

    aou will get:

     \displaystyle \int \frac {dx}{(2-x)^6} = -\int \frac {du}{u^6} = -\frac {1}{5(2-x)^5} + C


    P.S. sorry I didn't see that is just (2-x)^6 and sorry TheCoffeeMachine didn't see your post
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    Quote Originally Posted by bobsanchez View Post
    What is the method of integrating functions like 1/(1 + x^2)?
    For your original question, you have been told that this integral is \arctan{x} + C.

    The way they get this result is through trigonometric substitution. Trigonometric substitutions are handy to know, because through the use of the trigonometric identities, you can usually simplify the problem...

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

    Then

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

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

     = \int{\frac{\sec^2{\theta}}{\sec^2{\theta}}\,d\thet  a}

     = \int{1\,d\theta}

     = \theta + C

     = \arctan{x} + C.
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