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    I've forgotten how to integrate functions in the form of a fraction where the integral is not a log natural... can someone help me with this question?

    ∫ 2 / (x-2)^2
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    change it into the form 2(x-2)^-2

    then divide by the new power which would be -1, since the coefficient of x is 1 it has no effect

    dividing 2 by -1 gives -2

    so it becomes -2(x-2)^-1

    edit: you can also test the answer by differentiating it, so -1 x -2 = 2

    so therefore 2 x (x-2)^-2 = 2(x-2)^-2 which is the original function
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    (Original post by AndyOC)
    I've forgotten how to integrate functions in the form of a fraction where the integral is not a log natural... can someone help me with this question?

    ∫ 2 / (x-2)^2
    Rewrite that as \displaystyle 2 \int \frac{\mathrm{d}x}{(x-2)^2} = 2 \int (x-2)^{-2} \, \mathrm{d}x using the fact that \frac{1}{a^b} = a^{-b}.

    Then use the fact that \displaystyle \int x^n \, \mathrm{d}x = \frac{x^{n+1}}{n+1} + \mathcal{C}, what does that get you?
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    Oh, and moved to maths.
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    (Original post by C-rated)
    change it into the form 2(x-2)^-2

    then divide by the new power which would be -1, since the coefficient of x is 1 it has no effect

    dividing 2 by -1 gives -2

    so it becomes -2(x-2)^-1

    edit: you can also test the answer by differentiating it, so -1 x -2 = 2

    so therefore 2 x (x-2)^-2 = 2(x-2)^-2 which is the original function
    (Original post by Zacken)
    Rewrite that as \displaystyle 2 \int \frac{\mathrm{d}x}{(x-2)^2} = 2 \int (x-2)^{-2} \, \mathrm{d}x using the fact that \frac{1}{a^b} = a^{-b}.

    Then use the fact that \displaystyle \int x^n \, \mathrm{d}x = \frac{x^{n+1}}{n+1} + \mathcal{C}, what does that get you?
    got it thank you!
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    (Original post by Zacken)
    Rewrite that as \displaystyle 2 \int \frac{\mathrm{d}x}{(x-2)^2} = 2 \int (x-2)^{-2} \, \mathrm{d}x using the fact that \frac{1}{a^b} = a^{-b}.

    Then use the fact that \displaystyle \int x^n \, \mathrm{d}x = \frac{x^{n+1}}{n+1} + \mathcal{C}, what does that get you?
    what about this one? ∫ x / (4 - x^2)^1/2 dx
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    (Original post by AndyOC)
    what about this one? ∫ x / (4 - x^2)^1/2 dx
    Use the substitution u = 4-x^2.
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    thank you once again
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    (Original post by AndyOC)
    thank you once again
    No problem.
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    (Original post by Zacken)
    Use the substitution u = 4-x^2.
    Surely inspection is the easiest method here.
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    (Original post by B_9710)
    Surely inspection is the easiest method here.
    That's the way I'd do it as well, but he didn't spot the inspection so I suggested the sub. Which is fairly equivalent, really.
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    (Original post by Zacken)
    That's the way I'd do it as well, but he didn't spot the inspection so I suggested the sub. Which is fairly equivalent, really.
    In an exam it's best to do both methods to double check, you don't want to be losing marks when it comes to something as trivial as integration
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    (Original post by C-rated)
    In an exam it's best to do both methods to double check, you don't want to be losing marks when it comes to something as trivial as integration
    Agreed.
 
 
 
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