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    I'm not sure what method to use for this c4 integral:
    (x+1)^2/(x^2)+1

    Could you also tell me what method you're using and what type of integral you labelled this as? I've tried to multiply out the top part, recognition...but nothing seems to work


    thanks!!
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    (Original post by alpha_4)
    I'm not sure what method to use for this c4 integral:
    (x+1)^2/(x^2)+1

    Could you also tell me what method you're using and what type of integral you labelled this as? I've tried to multiply out the top part, recognition...but nothing seems to work


    thanks!!
    Note that \dfrac{(x+1)^2}{x^2+1} \equiv \dfrac{(x^2+1) + 2x}{x^2+1} and then recall that \dfrac{a+b}{c} \equiv \dfrac{a}{c} + \dfrac{b}{c}. This will leave you with two trivial integrals to deal with.
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    Try using long division first and see where that gets you

    EDIT: Sorry, why was this negged? What I said achieves exactly the same as what Farhan said - divide the top bit by the bottom bit (hence long division), to get an expression that's much simpler to integrate. I was just trying to help :facepalm:
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     \frac{(x+1)^2}{x^2+1} = \frac{x^2 +1 +2x}{x^2+1}
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    (Original post by Farhan.Hanif93)
    Note that \dfrac{(x+1)^2}{x^2+1} \equiv \dfrac{(x^2+1) + 2x}{x^2+1} and then recall that \dfrac{a+b}{c} \equiv \dfrac{a}{c} + \dfrac{b}{c}. This will leave you with two trivial integrals to deal with.
    thank you very much
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    (Original post by Farhan.Hanif93)
    Note that \dfrac{(x+1)^2}{x^2+1} \equiv \dfrac{(x^2+1) + 2x}{x^2+1} and then recall that \dfrac{a+b}{c} \equiv \dfrac{a}{c} + \dfrac{b}{c}. This will leave you with two trivial integrals to deal with.
    Sorry to bother you, but what would you do for this integral:
    1/(x^2)(x-1)

    thanks
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    (Original post by alpha_4)
    Sorry to bother you, but what would you do for this integral:
    1/(x^2)(x-1)

    thanks
    Use partial fractions.
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    (Original post by alpha_4)
    Sorry to bother you, but what would you do for this integral:
    1/(x^2)(x-1)

    thanks
    Partial fractions is probably a good idea.

    Express \dfrac{1}{x^2 (x-1)} in the form \dfrac{Ax+B}{x^2} + \dfrac{C}{x-1}, and find the constants A, B and C.
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    (Original post by Farhan.Hanif93)
    Partial fractions is probably a good idea.

    Express \dfrac{1}{x^2 (x-1)} in the form \dfrac{Ax+B}{x^2} + \dfrac{C}{x-1}, and find the constants A, B and C.
    How about this ?


    \frac 1{x^2(x-1)}=\frac {\large A}{x^2} + \frac {\large B}{x}+ \frac {\large C}{x-1}


    Oh wait. same thing. Sorry
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    (Original post by CookieGhoul)
    Oh wait. same thing. Sorry
    Yep.
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    (Original post by Farhan.Hanif93)
    Partial fractions is probably a good idea.

    Express \dfrac{1}{x^2 (x-1)} in the form \dfrac{Ax+B}{x^2} + \dfrac{C}{x-1}, and find the constants A, B and C.
    Thanks, I don't think those types of partial fractions are in the c4 syllabus but it seems like the most sensible option

    Only one more, if you don't mind...

    x^2/X^3 + 1
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    (Original post by alpha_4)
    Thanks, I don't think those types of partial fractions are in the c4 syllabus but it seems like the most sensible option

    Only one more, if you don't mind...

    x^2/X^3 + 1
    Is it (x^2)/(x^3+1) or (x^2)/(x^3) + 1
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    (Original post by alpha_4)
    Thanks, I don't think those types of partial fractions are in the c4 syllabus but it seems like the most sensible option

    Only one more, if you don't mind...

    x^2/X^3 + 1
    Note that \dfrac{x^2}{x^3+1} \equiv \dfrac{1}{3} \times \dfrac{3x^2}{x^3+1} \equiv \dfrac{1}{3} \times \dfrac{\frac{d}{dx}[x^3+C]}{x^3+1}. What do you know about integrals of that form. If that method doesn't work for you, use a substitution of u=x^3-1
    (Original post by CookieGhoul)
    Is it (x^2)/(x^3+1) or (x^2)/(x^3) + 1
    I'd assume that he meant the former as the latter is reasonably trivial.
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    (Original post by CookieGhoul)
    Is it (x^2)/(x^3+1) or (x^2)/(x^3) + 1
    (x^2)/(x^3) + 1
    the latter, sorry!
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    (Original post by Farhan.Hanif93)
    Note that \dfrac{x^2}{x^3+1} \equiv \dfrac{1}{3} \times \dfrac{3x^2}{x^3+1} \equiv \dfrac{1}{3} \times \dfrac{\frac{d}{dx}[x^3+C]}{x^3+1}. What do you know about integrals of that form. If that method doesn't work for you, use a substitution of u=x^3-1

    I'd assume that he meant the former as the latter is reasonably trivial.
    Didn't you just integrate the former?
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    (Original post by alpha_4)
    (x^2)/(x^3) + 1
    the latter, sorry!
    Omg dude. Then you'll be integrating

    \dfrac {1}{x} + 1
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    (Original post by alpha_4)
    (x^2)/(x^3) + 1
    the latter, sorry!
    You need to be using LaTeX to make sure your posts are clear. If you can't do that integral, you need to reopen your textbook and learn about integrals that lead to logs properly. Then come back and ask for help if you're stuck.
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    (Original post by Farhan.Hanif93)
    You need to be using LaTeX to make sure your posts are clear. If you can't do that integral, you need to reopen your textbook and learn about integrals that lead to logs properly. Then come back and ask for help if you're stuck.
    LOL! + rep to you

    Haha, you wrote so much for the other case, and he just wanted a simple integral
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    (Original post by CookieGhoul)
    LOL! + rep to you

    Haha, you wrote so much for the other case, and he just wanted a simple integral
    It's not so much that, it's just that I'm trying to help someone with a more difficult problem in the topic when he doesn't seem to understand the basics of it. :p:
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    LOOL, the difference latex would've made in this case is immense!
 
 
 
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