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    Sorry for another thread...I'm just trying to wrap up my learning for today

    Apparently if you integrate x/1+x² it equals 0.5ln(1+x²) + C

    Could someone show the step by step process of this for me, including saying what you're doing (ie if you are using a formula or using some kind of method)?

    It would reaaallllly help me!!
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    (Original post by Crazydavy)
    Sorry for another thread...I'm just trying to wrap up my learning for today

    Apparently if you integrate x/1+x² it equals 0.5ln(1+x²) + C

    Could someone show the step by step process of this for me, including saying what you're doing (ie if you are using a formula or using some kind of method)?

    It would reaaallllly help me!!
    Note that for  \dfrac{x}{1+x^2} the numerator is half the derivative of the denominator and then recall that \dfrac{d}{dx}[\mathrm{ln}f(x)] = \dfrac{f'(x)}{f(x)}
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    x/1+x² can be written as (1/2)*2x/1+x²

    There is a result that says the integral of f '(x)/f(x) = ln|f(x)| + c.

    So the integral of (1/2)*2x/1+x² is (1/2)*ln|1+x²| +c
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    (Original post by vc94)
    x/1+x² can be written as (1/2)*2x/1+x²

    There is a result that says the integral of f '(x)/f(x) = ln|f(x)| + c.

    So the integral of (1/2)*2x/1+x² is (1/2)*ln|1+x²| +c
    Thanks, I can see that working for this particular question but that's only because the numerator is conveniently the differential of the denominator when you split it apart - is there not another way to show it?
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    (Original post by Crazydavy)
    Thanks, I can see that working for this particular question but that's only because the numerator is conveniently the differential of the denominator when you split it apart - is there not another way to show it?
    Integration by substitution, put u= 1+x².
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    (Original post by Crazydavy)
    Thanks, I can see that working for this particular question but that's only because the numerator is conveniently the differential of the denominator when you split it apart - is there not another way to show it?
    THere is no need another way to show it. It is the most simple way. It is working for
    any so type question as your one because it is a general integration rule
    for this type of composite function. You should to recognize these types.
    It is fact that the numerator is the differential of the denominator. That is why
    you have to show that this intagral gives 1/2ln|1+x^2| +C.
    If there was only the 2 in the numerator you would to show that it gives 2arctanx+C.
    So when you integrate composite functions 'find the base function and the derivative as factor'.
    Without this maybe you can not give the integral agebrically.
    For example: \int 2x\cdot sin(x^2) dx is -cos(x^2)
    but \int sin(x^2) dx would be hard to give.
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    (Original post by Crazydavy)
    Thanks, I can see that working for this particular question but that's only because the numerator is conveniently the differential of the denominator when you split it apart - is there not another way to show it?
    Assuming this is an A-level question, what you have will always be either:
    (a) Something whose numerator is a constant multiple of the denominator
    (b) Something which can be split into partial fractions (perhaps after factorisation of the denominator)
    (c) Something which is a "function of a function" multiplied by the derivative of the "inside function" (e.g. \sin x \cos^5 x)
    (d) Something for which a substitution is given (C4) or obvious (FP2+)

    In this case you have (a), which is why it's so convenient :p:
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    x/ x^2 + 1 let u = x^2 + 1

    du/dx = 2x where dx = du / 2u

    integral ( x du / u 2x )

    simplifies

    integral (du / 2u) = 1/2 integral ( du /u ) = 1/2 ( u^-1 du )

    = 1/2 . ln (u) +C

    =1/2 . ln ( x^2 +1 ) + C
 
 
 

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