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 $\ln(f(x))$.