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#1 2005-08-11 19:41:52

fila
Novice

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tricky integration

can anyone help me come up with a solution to this
dy/dx=y/(a+y)

#2 2005-08-14 21:17:38

MathsIsFun
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Re: tricky integration

My integration skills leave a lot to be desired, but noone has attempted an answer yet, so I will have a go. At least someone can then pick holes in my poor attempts.

First of all, shouldn't that be dy/dx=x/(a+x) ?

Now, ∫ 1/x dx = ln x , and ∫ 1/f(x) dx = ln ( f(x) ) , so:
∫ x/(a+x) dx = ∫ 1 - a/(a+x) dx = ∫ dx - a ∫ 1/(a+x) dx = x - a ln(a+x) + C

Now I managed to find that ∫ ln(a+x) dx = a ln x + ln x
∫ x-a ln(a+x) + C dx = ∫ x dx - a ∫ ln(a+x) dx + C ∫ dx= x/2  -  a( a ln x -  ln x) + Cx + D

Can someone please check my work?


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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