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    Basically I don't understand anything. Any help would be so much appreciated or I'm going to fail...

    The one thats bugging me the most right now...Integration by parts of

     \frac{2x+4}{x^3-2x^2}

    I have tried it by making

     \frac{A}{x^2}   and   \frac{B}{x-2}


    and solving as a normal partial fraction to get A = =2 and B = 2 but these numbers don't seem to work and I don't know where I've gone wrong...
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    (Original post by fn_101)
    Basically I don't understand anything. Any help would be so much appreciated or I'm going to fail...

    The one thats bugging me the most right now...Integration by parts of

     \frac{2x+4}{x^3-2x^2}

    I have tried it by making

     \frac{A}{x^2}   and   \frac{B}{x-2}


    and solving as a normal partial fraction to get A = =2 and B = 2 but these numbers don't seem to work and I don't know where I've gone wrong...
    Your partial fractions are split wrong. When there is a repeated factor in the denominator you need to include it from power 1 all the way up to the power it is in the expression.

    I.e. in this case you need to use  \frac{A}{x}  and \frac{B}{x^2} and  \frac{C}{x-2}
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    (Original post by ukdragon37)
    Your partial fractions are split wrong. When there is a repeated factor in the denominator you need to include it from power 1 all the way up to the power it is in the expression.

    I.e. in this case you need to use  \frac{A}{x}  and \frac{B}{x^2} and  \frac{C}{x-2}

    Okay, so I tried it and got as far as following through this to get c=1 but i don't know what to put as the values of x to find A or B because they both have an x-2 ....
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    (Original post by fn_101)
    Basically I don't understand anything. Any help would be so much appreciated or I'm going to fail...

    The one thats bugging me the most right now...Integration by parts of

     \frac{2x+4}{x^3-2x^2}

    I have tried it by making

     \frac{A}{x^2}   and   \frac{B}{x-2}


    and solving as a normal partial fraction to get A = =2 and B = 2 but these numbers don't seem to work and I don't know where I've gone wrong...
    your factorising of the denominator is wrong. its a cubic function, so you've got to have 3 partial fractions
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    (Original post by fn_101)
    Okay, so I tried it and got as far as following through this to get c=1 but i don't know what to put as the values of x to find A or B because they both have an x-2 ....
    You need to expand out and compare the coefficients of constants, x, and x^2 etc. on left and right to get values for A and B.

    EDIT: Also check the c value you've already got.
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    (Original post by ukdragon37)
    You need to expand out and compare the coefficients of constants, x, and x^2 etc. on left and right to get values for A and B.

    EDIT: Also check the c value you've already got.
    Okay I have A= -2, B = -2 and C = 2....does that sound better to continue with?
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    (Original post by fn_101)
    Okay I have A= -2, B = -2 and C = 2....does that sound better to continue with?
    Yes.
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    (Original post by ukdragon37)
    Yes.
    Okay, right so using them it seems easier to intergrate but how do you intergrate

     \frac{-2}{x^2} ? I have used ln for the other functions (A and C) to get -2 ln x and 2 ln (x+2)
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    (Original post by fn_101)
    Okay, right so using them it seems easier to intergrate but how do you intergrate

     \frac{-2}{x^2} ? I have used ln for the other functions (A and C) to get -2 ln x and 2 ln (x+2)
    Surely that is easier to integrate than A and C? If you are doing this question then you should have done how to integrate  \frac{-2}{x^2} already.
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    I am also really stuck on these integrations..

     \frac{cos \sqrt{}x}{\sqrt {x}} using integration by substitution

    and

    if  y= tan^-1 x    and    \frac{dy}{dx} = \frac{1}{1 +x^2}

    Use inegration by parts to find the integral of

     x tan^-1 x dx
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    (Original post by fn_101)
    I am also really stuck on these integrations..

     \frac{cos \sqrt{}x}{\sqrt {x}} using integration by substitution
    We generally can deal with cosx, but cos(rootx) is harder. So the first substitution worth trying is u = sqrt(x)


    if  y= tan^-1 x    and    \frac{dy}{dx} = \frac{1}{1 +x^2}

    Use inegration by parts to find the integral of

     x tan^-1 x dx
    Let u = arctanx and dv/dx = x simply because the question hints at the differentiaton of arctanx. Although, in general, it's a good idea to let u = x (or whatever polynomial term).
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    (Original post by fn_101)
    Basically I don't understand anything. Any help would be so much appreciated or I'm going to fail...

    The one thats bugging me the most right now...Integration by parts of

     \frac{2x+4}{x^3-2x^2}

    I have tried it by making

     \frac{A}{x^2}   and   \frac{B}{x-2}


    and solving as a normal partial fraction to get A = =2 and B = 2 but these numbers don't seem to work and I don't know where I've gone wrong...

    http://www.wolframalpha.com/input/?i=integrate+(-2/x^2)

    wolframalpha was my saviour for a-level

    you can put in a problem and it solves it, even giving steps for the working.

    It's great to check answers/methods
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    (Original post by Swayum)
    We generally can deal with cosx, but cos(rootx) is harder. So the first substitution worth trying is u = sqrt(x)



    Let u = arctanx and dv/dx = x simply because the question hints at the differentiaton of arctanx. Although, in general, it's a good idea to let u = x (or whatever polynomial term).
    For the first one I tried...

     du = \frac {1} {2 \sqrt{x}}

    Then replaced the denominator and the dx, and took the 2 outside (or is it supposed to be 1/2 outside ? )

    Then ended up with the integral of 2 cos u which I can integrate


    For the second one I ended up with something I can't see myself integrating!

     tan^-1 x   .  \frac{x^2}{2} - the integral of \frac{x^2}{2} . \frac{1}{1+x^2}

    I dunno where I went wrong...
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    (Original post by fn_101)
    For the second one I ended up with something I can't see myself integrating!

     tan^-1 x   .  \frac{x^2}{2} - the integral of \frac{x^2}{2} . \frac{1}{1+x^2}

    I dunno where I went wrong...
    Multiply together the two fractions under the integral sign. Can you split up the fraction into a number and a fractional remainder?
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    (Original post by ukdragon37)
    Multiply together the two fractions under the integral sign. Can you split up the fraction into a number and a fractional remainder?
    You mean to get
     \frac {x^2}{(2)(1+x^2)}

    I'm confused..
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    (Original post by fn_101)
    You mean to get
     \frac {x^2}{(2)(1+x^2)}

    I'm confused..
    First note that

     \displaystyle\int \frac{x^2}{2\left(1+x^2\right)}d  x =\frac{1}{2}\displaystyle\int \frac{x^2}{\left(1+x^2\right)}dx

    Now can you perform some sort of polynomial long division (or otherwise) on the fraction in the integral to split it into "a whole number" + "a fraction in terms of x"?
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    (Original post by ukdragon37)
    First note that

     \displaystyle\int \frac{x^2}{2\left(1+x^2\right)}d  x =\frac{1}{2}\displaystyle\int \frac{x^2}{\left(1+x^2\right)}dx

    Now can you perform some sort of polynomial long division (or otherwise) on the fraction in the integral to split it into "a whole number" + "a fraction in terms of x"?
    Yes, I can Ahh thankyou!
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    Sorry to be a pain everyone but I do not understand limits at all.... for example, use the limit definition to show

     f(x)=((4+x^2)^2)

    then  f'(0) = 0
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    (Original post by fn_101)
    Sorry to be a pain everyone but I do not understand limits at all.... for example, use the limit definition to show

     f(x)=((4+x^2)^2)

    then  f'(0) = 0
    What level of maths are you at? What is your definition of a limit?

    And don't worry, you're not being a pain or anything. Keep em coming.
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    (Original post by fn_101)
    Sorry to be a pain everyone but I do not understand limits at all.... for example, use the limit definition to show

     f(x)=((4+x^2)^2)

    then  f'(0) = 0
    I suspect by "limit definition" you mean:
     f'(x) = \lim_{h \rightarrow 0} \dfrac{[4+(x+h)^2]^2-(4+x^2)^2}{h}

    The thing to remember is that you can only substitute h with 0 once it is not in the denominator. Since the question gives you a value of x, why don't you put that in and see what happens?
 
 
 
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