In my mind the integral of 1/x would be ln(x)+c not ln(x/c) ATH S FOR PHYSICS PR

Question

In my mind the integral of 1/x would be ln(x)+c not ln(x/c) ATH S FOR PHYSICS PROBLE S-SOLUTIONS can be integrat 1. Given pdyda+2, we see thst the ly with respect to whero the constant is written Aa /2 simply for t2P+A -general solution NB V-±ar+4x+1 or similar equally acoep Given dy/de(-1)/a, separate the variables, factorize -1-1)u+1 as follows: (b) y(+1) and apply partial fractions Integrating with respect to z leads to Now make y the subject B1/A2 for convenience and re-arrangin 1 + Ba 1- B / V generalsolution - 2(a). we see that dy dz ve 2'exact form If respect to z: 2…2 + Ap> -1+--general solution Or alternatively, by> g the variables 1 dy 2

Explanation / Answer

when you write constant C after an integral then it is not necessary to write it as positive C..It may be -ve also..

Now again if C is contant then log C will also be a constant so you can write it as log C also to make your calculations simple.

So integral of 1/x = ln x - ln c => ln x/c

How it is justified :

just differentiate ln x/c and we will get 1/x. Thus no matter how we are writing constant term. We can write it as per our choice to make our calculations simple. It doesnt matter its is -ve C, +ve C or Log C

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