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    Im doing P3, P1 and S1 next week.

    How do you integrate 8/ x^2 + 1 ????

    Can you integrate where the denominator has a power???

    Please help, thanx.
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    Is that 8 / (x^2 + 1) or (8 / x^2) + 1 ?
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    (Original post by BlueAngel)
    Im doing P3, P1 and S1 next week.

    How do you integrate 8/ x^2 + 1 ????

    Can you integrate where the denominator has a power???

    Please help, thanx.
    you need to specify what is the denominator.

    i suspect its (x^2 + 1) thats the total denominator (if it was just x^2, thats p1)

    there are several ways to do it, but the easiest is to look in the formula book other wise, some sort of substitution will do it.
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    (Original post by Morgan)
    Is that 8 / (x^2 + 1) or (8 / x^2) + 1 ?
    Its 8 / (x^2 + 1)

    Actually this is more difficult, its in the P3 exam paper.


    Evaluate this Limits are x = square root of 2 and x = 0

    Integral of x / squareroot of (16 - x^4) -


    Ive got the answers, just doesnt explain it.
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    (Original post by BlueAngel)
    Im doing P3, P1 and S1 next week.

    How do you integrate 8/ x^2 + 1 ????

    Can you integrate where the denominator has a power???

    Please help, thanx.

    Let x=tanu
    dx=sec^2u.du

    int.8/x^2+1.dx=8int.sec^2u/sec^2u.du=8u+C=8arctanx+C.
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    (Original post by BlueAngel)
    Its 8 / (x^2 + 1)

    Actually this is more difficult, its in the P3 exam paper.


    Evaluate this Limits are x = square root of 2 and x = 0

    Integral of x / squareroot of (16 - x^4) -


    Ive got the answers, just doesnt explain it.
    Okay you can just substitute and use u = x^2 +1 I think and glancing at the second one I think its a nominator-derivative of the denominator one.
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    (Original post by Ralfskini)
    Let x=tanu
    dx=sec^2u.du

    int.8/x^2+1.dx=8int.sec^2u/sec^2u.du=8u+C=8arctanx+C.

    OK RALFSKINI PLEASE TELL ME WHY YOU WOULD USE TAN HAS A SUBSTITUTION WHEN THE EQUATION INVOLVES ONLY NUMBERS, COS I AM SO CONFUSED NOW
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    (Original post by BlueAngel)
    Im doing P3, P1 and S1 next week.

    How do you integrate 8/ x^2 + 1 ????

    Can you integrate where the denominator has a power???

    Please help, thanx.
    The derivtative of tan^-1(x) is a standard form and is 1/(x^2 + 1).

    Using this, it's simply 8tan^-1x + C.

    The standard form is INT c/(ax^2 + b^2) = c.tan^-1(x/b)
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    (Original post by calumc)
    The derivtative of tan^-1(x) is a standard form and is 1/(x^2 + 1).

    Using this, it's simply 8tan^-1x + C.

    The standard form is INT c/(ax^2 + b^2) = c.tan^-1(x/b)



    Is this from P3??
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    (Original post by BlueAngel)
    OK RALFSKINI PLEASE TELL ME WHY YOU WOULD USE TAN HAS A SUBSTITUTION WHEN THE EQUATION INVOLVES ONLY NUMBERS, COS I AM SO CONFUSED NOW

    You can make this substitution because of the trig identity- 1+tan^x=sec^2x.
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    how do u guess wot to use for u in subtitution?
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    (Original post by BlueAngel)
    Is this from P3??
    I have no idea, I did Advanced Highers which are slightly different, and harder apparently.
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    (Original post by Ralfskini)
    Let x=tanu
    dx=sec^2u.du

    int.8/x^2+1.dx=8int.sec^2u/sec^2u.du=8u+C=8arctanx+C.
    being able to get arctan from an integration isn't on P3!!!

    Can someone please tell me how to do this question - cos it's getting me worried
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    (Original post by Silly Sally)
    being able to get arctan from an integration isn't on P3!!!

    Can someone please tell me how to do this question - cos it's getting me worried

    ..but I have only done up to P3. :confused:
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    (Original post by Ralfskini)
    ..but I have only done up to P3. :confused:
    Do you do edexcel or another board?
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    (Original post by Silly Sally)
    Do you do edexcel or another board?

    Edexcel.
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    (Original post by Ralfskini)
    Edexcel.
    AGHHHHH!! Same as me!!!! I can't do this integration. I would at first have involved an 'ln', like integral of 1/(3x + 1) = 1/3 ln (3x +1) <--- i think - now i don't know whther this is right or wrong!!!
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    (Original post by Silly Sally)
    AGHHHHH!! Same as me!!!! I can't do this integration. I would at first have involved an 'ln', like integral of 1/(3x + 1) = 1/3 ln (3x +1) <--- i think - now i don't know whther this is right or wrong!!!

 
 
 
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