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    I've hit a blank

    how to integrate  \int \frac{r}{r-2M} dr ?

    Many thanks.
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    Quotient rule
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    hint: r = r - 2M + 2M
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    (Original post by xfootiecrazeesarax)
    I've hit a blank

    how to integrate  \int \frac{r}{r-2M} dr ?

    Many thanks.
    Disclaimer: I've not touched A-level maths for two years now, but my first thought would be:

     \int \frac{r}{r-2M} dr = \int \left ( r \cdot \frac{1}{r-2M} \right ) dr

    Then use integration by parts
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    (Original post by xfootiecrazeesarax)
    I've hit a blank

    how to integrate  \int \frac{r}{r-2M} dr ?

    Many thanks.
    \frac{r}{r-2M} = \frac{r-2M + 2M}{r-2M} = 1 + \frac{2M}{r-2M}. The latter is then a standard logarithmic integral.
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    (Original post by Zacken)
    \frac{r}{r-2M} = \frac{r-2M + 2M}{r-2M} = 1 + \frac{2M}{r-2M}. The latter is then a standard logarithmic integral.
    that's the man i was looking for, thank you
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    (Original post by Uni12345678)
    Quotient rule
    Quotient rule is for differentiation, not integration
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    (Original post by Zacken)
    \frac{r}{r-2M} = \frac{r-2M + 2M}{r-2M} = 1 + \frac{2M}{r-2M}. The latter is then a standard logarithmic integral.
    This is why I hate integration. Just full of dumb arbitrary tricks like this. The weierstrass sub is probably the worst offender
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    (Original post by poonslayer9000)
    This is why I hate integration. Just full of dumb arbitrary tricks like this. The weierstrass sub is probably the worst offender
    This doesn't really make sense - it may seem arbitrary but it's a standard obvious technique to me. I'm sure loads of (non-integration) techniques you use are 'arbitrary and dumb' to someone else.

    And the weierstrass sub seems like the worse possible example in complaining about arbitrariness since it's one of the only systematic ways of solving any rational trigonometric integral. Emphasis on systematic.
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    (Original post by K-Man_PhysCheM)
    Quotient rule is for differentiation, not integration
    Oh sugar yes sorry haha
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    (Original post by xfootiecrazeesarax)
    I've hit a blank

    how to integrate  \int \frac{r}{r-2M} dr ?

    Many thanks.
    The standard approach is the substitution u=r-2M, which makes the fraction nicer, and doesn't introduce any new difficulties with the du since it's a linear transformation.

    I'd say that this is the approach to try first (i.e. look for a sub), as opposed to the cute "add 0" trick already suggested. Substitutions are generally applicable, but tricks are not.
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    (Original post by atsruser)
    The standard approach is the substitution u=r-2M, which makes the fraction nicer, and doesn't introduce any new difficulties with the du since it's a linear transformation.

    I'd say that this is the approach to try first (i.e. look for a sub), as opposed to the cute "add 0" trick already suggested. Substitutions are generally applicable, but tricks are not.
    The "add 0" trick is an algebraic simplification. Algebraic simplification to make an integral easier is certainly generally applicable.

    It's hard to say there's a definitive order of "you should try approach X before Y" with integrals, because there are always cases where it's fairly obvious you should do the opposite. That said, two heuristics I personally give high priority are:

    Is there a way to simplify this algebraically?
    Is there a way to simplify this using trig identities?

    I see many integrals posted on TSR where the poster would have benefited by exploring these options before jumping into integration specific methods.

    [That said, linear substitution is generally at least "benign" - it's unlikely to make things worse, even if it doesn't help].
 
 
 
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