Simple Indefinite Integral of \int\frac{4}{-e^{4x-7}} = ln(-e^{4x-7}) = -4x+7+C

[Procrastinate]

\int\frac{4}{-e^{4x-7}}

=ln-e^{4x-7}}

=-4x+7+C

The answer says it is:

=\-e^{-4x+7}+C

 

[D H]

The book is right, which is easily verifiable by computing the derivative the given answer.

 

[Procrastinate]

The book is right.

Is there somewhere I went wrong? I am guessing it has something to do with the ln.
I thought that

\ln{e}=1

Therefore: ln-e^{4x-7}}=-4x+7+C

 

[D H]

Yes, the problem is with the natural logarithm.

Just because \int \frac 1 x \,dx = \ln x does not mean that everything of the form \int \frac 1{f(x)}\,dx integrates to \ln(f(x)).