Thread: Order of integration

If I evaluate an integral
$\displaystyle \int {1 \over 3x}dx = \ln {3x} + \ln c $
or the constant can be brought outside: $\displaystyle \int {1 \over 3x}dx = {1 \over 3}\int {1 \over x} dx = {1 \over 3} \ln x + \ln c$
which are different answers, how is this?

Originally Posted by bigdoggy

If I evaluate an integral
$\displaystyle \int {1 \over 3x}dx = \ln {3x} + \ln c $
or the constant can be brought outside: $\displaystyle \int {1 \over 3x}dx = {1 \over 3}\int {1 \over x} dx = {1 \over 3} \ln x + \ln c$
which are different answers, how is this?

Would it be because:
$\displaystyle \ln 3x + \ln c = \ln 3 + \ln x + \ln c = \ln x + \ln c$
since the 3 is absorbed into the constant?

The reason I ask is because I've got this : $\displaystyle dy = - {du \over 5u} $ which can be solved to give different answers:
$\displaystyle 1) -y = \ln 5u + \ln c \rightarrow u = C \exp^{-y} \ 2)\ 5y = - \ln u - \ln c \rightarrow u = C \exp ^{-5y}$

how can this be so?

Not quite

int(dx/3x) is not ln(3x) but 1/3ln(3x) +c = 1/3ln(x) + c (1/3ln(3)is absorbed into c

To see this differentiate ln(3x) -- you don't get 1/3x

When integrating int(dx/3x) if you leave the 3 in the integral

you should use substitution u = 3x du = 3dx

1/3du = dx and you get 1/3int(1/udu)