Logic behind the Cs in Integration

[Jason76]

Some people right this differently (ex: with no C on the left side equation from the start, with no attempt to get rid of Cs). But what's the logic behind C coming on both sides of the equation, but they only = 1 C.

\(\displaystyle \int \dfrac{1}{y + 1} dy = \int \dfrac{1}{6x + 9} dx\)

\(\displaystyle \ln (y + 1) + C_{a} = \ln (6x + 9) + C_{b}\)

\(\displaystyle \ln (y + 1) + C_{a} - C_{a} = \ln (6x + 9) + C_{b} - C_{a}\)

\(\displaystyle \ln (y + 1) = \ln (6x + 9) + C\)

 

[tkhunny]

You may track different constants if you like. In the end, though, addition still works.

 

[Jason76]

You may track different constants if you like. In the end, though, addition still works.

What do you mean by addition?

 

[tkhunny]

An arbitrary constant on each side of an equation is no more or less arbitrary than a single arbitrary constant on only one side.

 

[HallsofIvy]

Some people right this differently (ex: with no C on the left side equation from the start, with no attempt to get rid of Cs). But what's the logic behind C coming on both sides of the equation, but they only = 1 C.

\(\displaystyle \int \dfrac{1}{y + 1} dy = \int \dfrac{1}{6x + 9} dx\)

\(\displaystyle \ln (y + 1) + C_{a} = \ln (6x + 9) + C_{b}\)

\(\displaystyle \ln (y + 1) + C_{a} - C_{a} = \ln (6x + 9) + C_{b} - C_{a}\)

\(\displaystyle \ln (y + 1) = \ln (6x + 9) + C\)

The "logic" is just what you did here. Taking a "C" on each side, you can subtract them. And since C_b and C_a are arbitrary, C_b- C_a is an arbitrary number that you can just call "C".