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# By parts Int is pants!!!!!!!!!! I(1/(x+1)) watch

1. Ok i keep using parts and this is pants

im trying to integrate 1/(x+1) how can i do this???

The Q. was dy/dx - y/(x+1) = x

i got it down to

d/dx(y(x+1)^-1) = x/(x+1) now im trying to integrate the RHS
2. It's just ln(x+1) + C.

d[ln(ax+b)]/dx = a/(ax+b)
=> int 1/(ax+b) dx = (1/a) ln(ax+b) + C

And: "pants"?
3. (Original post by dvs)
It's just ln(x+1) + C.

d[ln(ax+b)]/dx = a/(ax+b)
=> int 1/(ax+b) dx = (1/a) ln(ax+b) + C

And: "pants"?

oooops i meant x/(x+1) sowwy
4. x/(x+1) = (x+1-1)/(x+1) = (x+1)/(x+1) - 1/(x+1) = 1 - 1/(x+1)
5. (Original post by dvs)
x/(x+1) = (x+1-1)/(x+1) = (x+1)/(x+1) - 1/(x+1) = 1 - 1/(x+1)
therefore the integral is

x-ln(x+1) ja?
6. (Original post by futureaussiecto)
therefore the integral is

x-ln(x+1) ja?
Yes
7. Why would you need parts for this?

As far as I'm concerned you only need parts if youre integrating a product of a polynomial AND ONE OF a trig, exponential or ln function.
8. (Original post by distortedgav)
Why would you need parts for this?

As far as I'm concerned you only need parts if youre integrating a product of a polynomial AND ONE OF a trig, exponential or ln function.
It has other applications, e.g. for finding the integral of (e^x)(sin(x)) or (e^x)(cos(x)) as well as finding reduction formulae.

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Updated: February 23, 2006
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