Intergral from P1 to P2 of P/(P+b) dP where b is a constant. I'll code it cleaner soon

Question

anonymous · Answer

$$\int\limits_{P_1}^{P_2}{P \over P+b}dP$$

turingtest · Answer

try adding b-b in the numerator

anonymous · Answer

So that would leave me with -b over P+b, then solve

turingtest · Answer

1-b/(P+b)

turingtest · Answer

then integrate

anonymous · Answer

I would still have to do u-substitution for the denominator. And you're right there should be a 1 infront, thx

turingtest · Answer

yes, but the u-sub in the denominator is trivial, because its differential is already dP

anonymous · Answer

$$P_2-P_1-b \ln {P_2+b \over P_1+b}$$

turingtest · Answer

without breaking out pen and paper, I think that's right

anonymous · Answer

Thanks for your help. I forget that a problem can be manipulated by adding in "0"

turingtest · Answer

yeah, that's right
and yeah, gotta love that +a-a trick, otherwise you're stuck doing polynomial long division... yuck