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    What's the way to to go about it?

    the question was to find the integration of (x^2 + 5) / (x^2 + 9). Dividing the numer. by denom. gives integration of (1 - 4/(x^2 + 9)). Now what?
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    Make the substitution x=3\tan \theta and use the identity 1 + \tan^2 \theta \equiv \sec^2 \theta.

    [Or, if you're lazy, look in the formula book.]
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    so you've got 1-(4/x^2 + 9)
    integrating 1 gets you x, so now you have -4/(x^2 + 9)
    to integrate
    treat it as 1/(x^2 + 9) this is (1/2) ln(x^2 + 9)
    then times it by the -4, to get -2 ln(x^2 + 9)

    so overall you get x - 2 ln(x^2 + 9)

    sorry if this isn't very clear
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    (Original post by proud nd luvin it)
    so you've got 1-(4/x^2 + 9)
    integrating 1 gets you x, so now you have -4/(x^2 + 9)
    to integrate
    treat it as 1/(x^2 + 9) this is (1/2) ln(x^2 + 9)
    then times it by the -4, to get -2 ln(x^2 + 9)

    so overall you get x - 2 ln(x^2 + 9)

    sorry if this isn't very clear
    Isn't integ. of (1/something) = (ln something) just for linear expressions?
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    (Original post by proud nd luvin it)
    so you've got 1-(4/x^2 + 9)
    integrating 1 gets you x, so now you have -4/(x^2 + 9)
    to integrate
    treat it as 1/(x^2 + 9) this is (1/2) ln(x^2 + 9)
    then times it by the -4, to get -2 ln(x^2 + 9)

    so overall you get x - 2 ln(x^2 + 9)

    sorry if this isn't very clear
    int [f'(x)/f(x)] = ln(f(x))

    what you have said here is clearly wrong.

    use the tan substitution OP
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    (Original post by nuodai)
    Make the substitution x=3\tan \theta and use the identity 1 + \tan^2 \theta \equiv \sec^2 \theta.

    [Or, if you're lazy, look in the formula book.]
    But why would x be equal to 3tan(theta)? Because the final answer is given in terms of x
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    (Original post by gunmetalpanda)
    Isn't integ. of (1/something) = (ln something) just for linear expressions?

    (Original post by slylion1)
    int [f'(x)/f(x)] = ln(f(x))

    what you have said here is clearly wrong.

    use the tan substitution OP
    Oh crap... yes you're right sorry but still... how come you'd use the tan substitution??
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    (Original post by gunmetalpanda)
    But why would x be equal to 3tan(theta)? Because the final answer is given in terms of x
    .
    You substitute x=3\tan \theta, integrate w.r.t. \theta, and then make the reverse substitution \theta = \tan^{-1} \dfrac{x}{3} to put your answer in terms of x again. You substitute x=3\tan \theta because then the 3 squares to a 9, which you can take out as a common factor, and then you can apply the identity on the denominator of the fraction. Because you get dx = \sec^2 \theta\, d\theta, the \sec^2 \theta terms end up cancelling and you get something simple to integrate.
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    Also, could i just say that
    Spoiler:
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     \displaystyle\int\dfrac{1}{x^2+a  ^2} = \dfrac{1}{a}tan^{-1}(\dfrac{x}{a})+C


    So in this case, a would be 3.


    _Kar.
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    (Original post by Kareir)
    Also, could i just say that

    Spoiler:
    Show
     \displaystyle\int\dfrac{1}{x^2+a  ^2} = \dfrac{1}{a}tan^{-1}(\dfrac{x}{a})+C


    So in this case, a would be 3.


    _Kar.
    Yes (but that's cheating [I mean not actually cheating, but it takes the fun out of integration]).
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    Haha, i've spoilered it, just to keep a bit of the mystery in

    _Kar.
 
 
 
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