Using partial fractions, we can rewrite the equation as 1 dN = dt rN 50000r - rN dN Or rN 50000r –rN dN = dt (i) Comment Step 2 of 4 A Integrate both sides, 1 dN = [dt +C N 50000r – rN - In |50000r – rN| =1+C

Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter7: Applying Fractions
Section7.3: Equations With Fractional Coefficient
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i don't understand these steps. please show all the steps to integrate this function

I'm confused about:
1) for the first term if 1/r can be taken as a common constant, also it can be done with the second term too, then why there is 'r' in the ln function still after taking 1/r as a common factor?

2) how did the sign change? 

Using partial fractions, we can rewrite the equation as
1
1
dN = dt
rN 50000r –rN ,
dN
Or
rN 50000r – rN
dN
= dt
(i)
Comment
Step 2 of 4 A
Integrate both sides,
dN
1
-dN = [dt + C
N
50000r – rN
Transcribed Image Text:Using partial fractions, we can rewrite the equation as 1 1 dN = dt rN 50000r –rN , dN Or rN 50000r – rN dN = dt (i) Comment Step 2 of 4 A Integrate both sides, dN 1 -dN = [dt + C N 50000r – rN
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