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Integration Help

k;
(edited 13 years ago)
Reply 1
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nuodai
17
Well how else can you write u+1u\dfrac{u+1}{u}?
Reply 2
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Kareir
2
Original post by hisname
How would i integrate x/(x-1)?

I've taken a substitution of u=x-1 and re-written the expression as

(U+1)/U but i don't know what do next i'm guessing its some sort of natural log but i'm not sure.

Thanks in advance for your help.



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_Kar.
Reply 3
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hisname
OP
';';
(edited 13 years ago)
Reply 4
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hisname
OP
Original post by nuodai
Well how else can you write u+1u\dfrac{u+1}{u}?


How exactly would i split the fraction like this

x* 1(x-1) and do integration by parts.
Original post by hisname
How exactly would i split the fraction like this

x* 1(x-1) and do integration by parts.


You're creating too much work for yourself.

Think about what fractions adding and subtracting actually does.

I.e, x/4 + 2y/4 = x+2y/4

You can then split it up again into the same fractions.

When two coefficients share a denominator by the action of addition or subtraction, you can split them up simply by putting the sign in the middle and putting them over separate denominators of the same value.

Think about what you would get in this case.
Reply 6
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hisname
OP
Is this right? you can re-write (U+1)/U as (U/U)+(1/u) or (1+(1/u))

On integrating that gives u+ ln u +c
Reply 7
Avatar for nuodai
nuodai
17
Original post by hisname
Is this right? you can re-write (U+1)/U as (U/U)+(1/u) or (1+(1/u))

On integrating that gives u+ ln u +c


Yup, and then you can substitute back to get it in terms of x.
Reply 8
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hisname
OP
Thank you very much for all your help everyone.
no you need to differentiate u with respect to x because of the product rule, and multiply it by the function where u is substituted for x to find your integrand
(edited 13 years ago)
Reply 10
Avatar for nuodai
nuodai
17
Original post by thievingllama
no you need to differentiate u with respect to x because of the product rule, and multiply it by the function where u is substituted for x to find your integrand


The product rule definitely doesn't come into this, but if you're referring to the part in integration by substitution where you multiply by du/dx, it doesn't matter in this case since it evaluates to 1.

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