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Indefinite integral error

So I did this indifinite integral one way and got the wrong answer. Then I did it another way (right side), and got the correct answer. Still, I can't figure out why the first approach didn't work. Would appreciate any explanation!

Attachment 21840

Re: Indefinite integral error

Both methods are correct. Your mistake is in the third line, you just made a mistake in your integration.

$\displaystyle 2\int \frac{-e^{-x}}{e^{-x}+1}dx=2ln(e^{x}+1)-2x + C$

With this correction, your other 2x vanishes and you are left with a similar expression.

Re: Indefinite integral error

If you write:

2x+2ln(1+e^(-x))+C

as

2(x+ln(1+e^(-x))+C

=2(ln(e^x)+ln(1+e^(-x)))+C

=2(ln((e^x).(1+e^(-x)))+C

=2 ln(e^x+1)+C

Re: Indefinite integral error

$\displaystyle \frac{2}{e^{-x} + 1} = \frac{2e^x}{1+e^x}$

That help?

Re: Indefinite integral error

@ Gralla55:

There's nothing wrong with your solutions.

Also it's not necessary to use 'absolute value signs' because e^(-x)+1>0 for each x.

@DivisionByZero:

That's not a mistake because:

2ln(e^x+1)-2x=2ln(e^x+1)-2ln(e^x)=2ln((e^x+1)/(e^x))=2ln(e^(-x)+1)

So just the same as Gralla55 said.