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The solution book gives this as an answer:

$\int \frac{1-e^x}{e^x}dx = \int \frac{1}{e^x}-\frac{e^x}{e^x}dx = -e^{-x} + C$

I would think it would be solved this way:

$\int \frac{1-e^x}{e^x} dx = \int \frac{1}{e^x} -1 \; dx = -e^{-x}-x + C$

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    @J.D. This is book was to be a brief review of calculus and in every section there is a problem that either can't be solved with basic methods described like the integral of sin(x^2), or the solutions given are just wrong. I am getting good at finding these errors, I might just do this for a job one day. I bought a terrible book to review, but there are not many to choose from around here.2012-04-01

2 Answers 2

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Yes, your approach is correct.

But, remember, you can always check your answer when finding an antiderivative:

We have: $ {d\over dx} (-e^{-x}+C)=e^{-x}\ne e^{-x}-1={1-e^x\over e^x}; $ so the book is wrong.

However:

$ {d\over dx} (-e^{-x}-x+C)=e^{-x}-1 = {1-e^x\over e^x}; $ so, you are right.

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    @David Mitra tha$n$ks for giving it a better title.2012-03-31
2

The best way to see if you been right in solving your integral is to derive both sides.

As we can see, the books answer:

$\frac{{d}(-e^{-x}+C)}{dx}\neq \frac{ d\{{\int \frac{1}{e^x}-\frac{e^x}{e^x}dx}\}}{dx} $