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    I've been doing some differential equations, but I'm having a bit of trouble with some if the integration involved

    How do you integrate these?:

    1/(square root of) 1+ 2y
    1/-2(square root of)y

    Thank you!
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    (Original post by EmJ15)
    I've been doing some differential equations, but I'm having a bit of trouble with some if the integration involved

    How do you integrate these?:

    1/(square root of) 1+ 2y
    1/-2(square root of)y

    Thank you!
    Indices rule:

    The first one is \displaystyle \int \frac{1}{\sqrt{1 + 2y}}\ , \mathrm{d}y= \int \frac{1}{(1 + 2y)^{1/2}} \, \mathrm{d}y=\int (1 + 2y)^{-1/2} \, \mathrm{d}y now use the 'reverse chain rule' on that.

    The second one is \displaystyle \int -\frac{1}{2\sqrt{y}} \, \mathrm{d}y = -\frac{1}{2} \int y^{-1/2} \, \mathrm{d}y then do your usual add one to the power and divide by the new power.
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    Well, since square roots are involved, I imagine you'd be looking at using a substitution along the lines of u = 1+ 2y! Try that and see if it gets you anywhere

    EDIT: Okay Zacken is here never mind he's right haha
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    The first one can be done by inspection- note that  \dfrac{d}{dy}(1+2y)^{ \frac{1}{2}} = 0.5 \times 2(1+2y)^{-0.5} which is the required integral
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    (Original post by Zacken)
    Indices rule:

    The first one is \displaystyle \int \frac{1}{\sqrt{1 + 2y}}\ , \mathrm{d}y= \int \frac{1}{(1 + 2y)^{1/2}} \, \mathrm{d}y=\int (1 + 2y)^{-1/2} \, \mathrm{d}y now use the 'reverse chain rule' on that.

    The second one is \displaystyle \int -\frac{1}{2\sqrt{y}} \, \mathrm{d}y = -\frac{1}{2} \int y^{-1/2} \, \mathrm{d}y then do your usual add one to the power and divide by the new power.
    (Original post by Don Pedro K.)
    Well, since square roots are involved, I imagine you'd be looking at using a substitution along the lines of u = 1+ 2y! Try that and see if it gets you anywhere

    EDIT: Okay Zacken is here never mind he's right haha
    ffs I literally cant type fast enough XD
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    (Original post by Zacken)
    Indices rule:

    The first one is \displaystyle \int \frac{1}{\sqrt{1 + 2y}}\ , \mathrm{d}y= \int \frac{1}{(1 + 2y)^{1/2}} \, \mathrm{d}y=\int (1 + 2y)^{-1/2} \, \mathrm{d}y now use the 'reverse chain rule' on that.

    The second one is \displaystyle \int -\frac{1}{2\sqrt{y}} \, \mathrm{d}y = -\frac{1}{2} \int y^{-1/2} \, \mathrm{d}y then do your usual add one to the power and divide by the new power.
    Thank so much, that has helped loads. Could I just ask you 1 question though, what is reverse chain rule and is there a particular scenario when you use it? Sorry, it's just that I don't think we've covered it in school. It's not in the book we use anyway. Thank you!
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    (Original post by Don Pedro K.)
    Well, since square roots are involved, I imagine you'd be looking at using a substitution along the lines of u = 1+ 2y! Try that and see if it gets you anywhere

    EDIT: Okay Zacken is here never mind he's right haha
    Haha, thanks anyway!
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    (Original post by EmJ15)
    Thank so much, that has helped loads. Could I just ask you 1 question though, what is reverse chain rule and is there a particular scenario when you use it? Sorry, it's just that I don't think we've covered it in school. It's not in the book we use anyway. Thank you!
    It's basically this:
    \displaystyle

\begin{equation*} \int (ax+b)^n \, \mathrm{d}x = \frac{(ax+b)^{n+1}}{a(n+1)} + \mathcal{C} \end{equation*}
 
 
 
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