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    Hi can anyone help me with this integration question thanks so much

    integrate x^2/(1+ x^2) dx
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    Substitution?
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    Everyone seems to be asking me how to do that integral these days for some reason

    Anyway, I think the advice I gave everyone else was to use this substitution:

    x = tan(u)

    and then you'll end up having to integrate tan^2(u) du

    but remember, tan^2(u) = sec^2(u) - 1, and sec^2(u) integrates to give tan (u)


    I think the answer is something like x - arctan(x) + c - Just check that
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    \dfrac{x^2}{1+x^2} = \dfrac{x^2 + 1 - 1}{x^2 + 1} = 1 - \dfrac{1}{x^2+1}.
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    (Original post by caaakeeey)
    Substitution?
    i tried both u = 1 + x^2 and u = x^2 didn't work for me
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    When the degree of the numerator is greater than or equal to the degree of the denominator, you want to start by dividing the fraction through (this is essentially what Acid has done).

    By degree we mean the highest power of x (and it's only really relevant when dealing with polynomials).
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    thanks for your help Glutamic Acid and tazarooni89!
    and also caaakeeey
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    (Original post by tazarooni89)
    Everyone seems to be asking me how to do that integral these days for some reason

    Anyway, I think the advice I gave everyone else was to use this substitution:

    x = tan(u)

    and then you'll end up having to integrate tan^2(u) du

    but remember, tan^2(u) = sec^2(u) - 1, and sec^2(u) integrates to give tan (u)

    I think the answer is something like x - arctan(x) + c - Just check that
    Yes, I see what you mean

    but I think the answer is
    x - tan ^-1(x) = x - arctan(x)
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    (Original post by Stricof)
    Yes, I see what you mean

    but I think the answer is x - tan ^-1(x)
    Isn't that what I said?

    x - arctan(x) + c

    c is the constant of integration remember :p:
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    (Original post by tazarooni89)
    Isn't that what I said?

    x - arctan(x) + c

    c is the constant of integration remember :p:
    Mmm...yes...oh bugger your right.
    Also i edited my answer lol
    Sorry man

    I got x - arctan(x)
    Guess...its' the same LOL!
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    (Original post by Swayum)
    When the degree of the numerator is greater than or equal to the degree of the denominator, you want to start by dividing the fraction through (this is essentially what Acid has done).

    By degree we mean the highest power of x (and it's only really relevant when dealing with polynomials).
    got it :yep: thanks
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    Can't you just use the quotient rule?

    (1+x^2)*(2x) - x^2*2x / (1+ x^2)^2

    (2x^2x) - (2x^3) / (1+ x^2)^2

    2x / (1+ x^2)^2
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    (Original post by Judas the Prude)
    Can't you just use the quotient rule?

    (1+x^2)*(2x) - x^2*2x / (1+ x^2)^2

    (2x^2x) - (2x^3) / (1+ x^2)^2

    2x / (1+ x^2)^2
    No. That's for differentiation.
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    (Original post by Judas the Prude)
    Can't you just use the quotient rule?

    (1+x^2)*(2x) - x^2*2x / (1+ x^2)^2

    (2x^2x) - (2x^3) / (1+ x^2)^2

    2x / (1+ x^2)^2
    Yes, if the question was to differentiate it. But we're integrating here :p:
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    (Original post by Glutamic Acid)
    No. That's for differentiation.

    Oh. LOL. I wondered why no one else had suggested it. Fail.

    Thanks.
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    (Original post by Glutamic Acid)
    No. That's for differentiation.
    Lol yes thats true...
    for no reason but stupidity I differentiated the equation of the OP

    got
    (-2*x^3)/(1 + x^2)^2 + (2*x)/(1 + x^2)

    I think this is wrong


    I know this has no relevance to the actual question, but...just wondering
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    (Original post by Stricof)
    Lol yes thats true...
    for no reason but stupidity I differentiated the equation of the OP

    got
    (-2*x^3)/(1 + x^2)^2 + (2*x)/(1 + x^2)

    I think this is wrong


    I know this has no relevance to the actual question, but...just wondering
    It's not wrong, but it can be simplified to 2x/(1+x^2)^2

    Or if you dont like brackets: 2x/x^4 + 2x^2 + 1 :yep:



    If you wanted to differentiate it easily, I'd suggest looking at post #4 first :p:
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    (Original post by tazarooni89)
    It's not wrong, but it can be simplified to 2x/(1+x^2)^2 :yep:
    Oh yes...damn, just like the post above my original answer

    Okay that was a fail on my accord as well lol
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    (Original post by tazarooni89)
    Everyone seems to be asking me how to do that integral these days for some reason

    Anyway, I think the advice I gave everyone else was to use this substitution:

    x = tan(u)

    and then you'll end up having to integrate tan^2(u) du

    but remember, tan^2(u) = sec^2(u) - 1, and sec^2(u) integrates to give tan (u)


    I think the answer is something like x - arctan(x) + c - Just check that
    But does it not become 1/(1+x^2) = 1/(1+ tan^2 u) = 1/ sec^2 u = cos^2 u?

    Thanks
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    (Original post by getreadygetset)
    But does it not become 1/(1+x^2) = 1/(1+ tan^2 u) = 1/ sec^2 u = cos^2 u?

    Thanks
    No, because when you use the substitution x = tan u
    you also have to remember to say that dx = sec^2(u) du
 
 
 
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