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    Find the integral of 1/[1+(x-1)^1/2] using substitution u^2 = x-1

    I'm sure this should be really easy, but I get to

    2u/1+u

    which I'm unable to integrate



    Edit: Rep available!
    • Thread Starter
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    Partial fractions? But there's only one denominator, which doesn't factorise -how do I do that?

    Feeling stupid(er) now
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    Fine, delete your post and make me look worse :p: thanks for that
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    think by parts... make it 2u(1+u)-1
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    (Original post by Apagg)
    Find the integral of 1/[1+(x-1)^1/2] using substitution u^2 = x-1

    I'm sure this should be really easy, but I get to

    2u/1+u

    which I'm unable to integrate
    2u/1+u = (2+2u-2)/1+u = 2 - 2/(1+u), and then integrate as usual.
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    (Original post by Axon)
    2u/1+u = (2+2u-2)/1+u = 2 - 2/(1+u), and then integrate as usual.
    nice... even better than by parts! hehe
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    You genius Axon! Thank you, rep coming your way

    Hate those sort of questions
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    (Original post by Apagg)
    You genius Axon! Thank you, rep coming your way

    Hate those sort of questions
    you are welcome anyway!
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    Sorry But the question wants substitution only, so I couldn't use your method. Plus, you made me look silly :p:
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    (Original post by Apagg)
    Sorry But the question wants substitution only, so I couldn't use your method. Plus, you made me look silly :p:
    well u can use any method u want... because u DID use the substitution! (when did i make u look silly?)
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    You made a post and then deleted it, so my response to the post made it seem I was going crazy

    I'm working through the book step by step, so only using the methods that have cropped up up to that point, if that makes sense
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    (Original post by Apagg)
    You made a post and then deleted it, so my response to the post made it seem I was going crazy

    I'm working through the book step by step, so only using the methods that have cropped up up to that point, if that makes sense
    haha.. well i deleted it because i didnt see that u already reached where i stopped! sorry anyway...
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    Not to worry
 
 
 
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