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    The answer at the back of the book is different to what I got:|



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    Oops forgot a pic!


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    (Original post by livealittle)
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    Oops forgot a pic!


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    Everything you've written down seems to be correct. By any chance, does the book not include the 2? If that's the case, they've just added it to the constant of integration since 2 is a constant term.

    \displaystyle \begin{aligned} \int \frac{dx}{1+\sqrt{x}} & = 2 + 2\sqrt{x} - 2\ln \left( 1+ \sqrt{x} \right) + \zeta \\ & = 2\sqrt{x} - 2\ln \left( 1+ \sqrt{x} \right) + \zeta \end{aligned}
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    (Original post by livealittle)
    The answer at the back of the book is different to what I got:|



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    What does the book say
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    (Original post by Khallil)
    Everything you've written down seems to be correct. By any chance, does the book not include the 2? If that's the case, they've just added it to the constant of integration since 2 is a constant term.

    \displaystyle \begin{aligned} \int \frac{dx}{1+\sqrt{x}} & = 2 + 2\sqrt{x} - 2\ln \left( 1+ \sqrt{x} \right) + \zeta \\ & = 2\sqrt{x} - 2\ln \left( 1+ \sqrt{x} \right) + \zeta \end{aligned}
    Yeah the book doesn't include the 2.ohh so it is because 2 is not a variable of x?
    I think I get it now ,Thanks
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    (Original post by TenOfThem)
    What does the book say
    2 x^1/2 - 2ln( 1+x^1/2)


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    (Original post by livealittle)
    Yeah the book doesn't include the 2.ohh so it is because 2 is not a variable of x?
    I think I get it now ,Thanks
    Yep. 2 is a constant and doesn't depend on any value of x. As a result we can incorporate it into the constant of integration which also doesn't depend on x.

    Here's a substitution that'll take you straight to the answer without the 2
    \text{Let} \ u = \sqrt{x}
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    (Original post by livealittle)
    2 x^1/2 - 2ln( 1+x^1/2)


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    (Original post by Khallil)
    Yep. 2 is a constant and doesn't depend on any value of x. As a result we can incorporate it into the constant of integration which also doesn't depend on x.

    Here's a substitution that'll take you straight to the answer without the 2
    \text{Let} \ u = \sqrt{x}
    Is zeta commonly used instead of C to denote the integration constant?
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    (Original post by TenOfThem)
    No +c
    Haha! I always forget the constant !


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    (Original post by Khallil)
    Yep. 2 is a constant and doesn't depend on any value of x. As a result we can incorporate it into the constant of integration which also doesn't depend on x.

    Here's a substitution that'll take you straight to the answer without the 2
    \text{Let} \ u = \sqrt{x}
    Thanks mate really helpful


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    (Original post by keromedic)
    Is zeta a commonly used instead of C to denote the integration constant?
    Nope, I just got bored of the letter c. :sleep:

    (Original post by livealittle)
    Thanks mate really helpful
    You're welcome, glad you got it!
 
 
 
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