e Integration Problem

[Jason76]

\(\displaystyle \int \dfrac{dx}{1 + e^{x}}\) use formula \(\displaystyle \int \dfrac{du}{u} = \ln u + C\)

\(\displaystyle \int \dfrac{e^{x} dx}{1 + e^{x}}\) Make the dx match the du. The \(\displaystyle y'\) of \(\displaystyle e^{x}\) is \(\displaystyle e^{x}\) so du = \(\displaystyle e^{x}\)

\(\displaystyle \dfrac{1}{e^{x}} \int \dfrac{e^{x} dx}{1 + e^{x}} = \int \dfrac{1}{e^{x}} * \ln (1 + e^{x}) + C\) Balance the equations with a reciprocal. Final Answer

Book shows \(\displaystyle - \int \dfrac{-e^{-x} dx}{e^{-x} + 1} = -\ln(1 + e^{-x}) + C\) as the answer. ??

 

[MarkFL]

We are given to evaluate:

\(\displaystyle \displaystyle \int\frac{dx}{1+e^x}\)

I would multiply the integrand by \(\displaystyle 1=\dfrac{e^{-x}}{e^{-x}}\) and write the integral as:

\(\displaystyle \displaystyle -\int\frac{-e^x}{e^{-x}+1}\,dx\)

Now, use the substitution:

\(\displaystyle u=e^{-x}+1\,\therefore\,du=-e^{-x}\,dx\) and we have:

\(\displaystyle \displaystyle -\int\frac{du}{u}=-\ln|u|+C=-\ln\left(e^{-x}+1 \right)+C\)

Does this make sense?

 

[Jason76]

We are given to evaluate:

\(\displaystyle \displaystyle \int\frac{dx}{1+e^x}\)

I would multiply the integrand by \(\displaystyle 1=\dfrac{e^{-x}}{e^{-x}}\) and write the integral as:

\(\displaystyle \displaystyle -\int\frac{-e^x}{e^{-x}+1}\,dx\)

Now, use the substitution:

\(\displaystyle u=e^{-x}+1\,\therefore\,du=-e^{-x}\,dx\) and we have:

\(\displaystyle \displaystyle -\int\frac{du}{u}=-\ln|u|+C=-\ln\left(e^{-x}+1 \right)+C\)

Does this make sense?

Is your answer and mine the same answer in different forms?

 

[MarkFL]

My answer is the same as what your book gives.

Your working still has an integral in the final answer, and you have illegally put a function of x outside of the integral to balance a function put inside. You can only do this with constants.

 

[Deleted member 4993]

Another way:

\(\displaystyle \displaystyle \int\frac{dx}{1+e^x}\)

substitute:

u = e^x

du = e^x dx

dx = du/u

\(\displaystyle \displaystyle \int\frac{dx}{1+e^x}\)

\(\displaystyle = \ \displaystyle \int\frac{du}{u(1+u)}\)

\(\displaystyle = \ \displaystyle \int\frac{du}{u} - \ \displaystyle \int\frac{du}{(1+u)}\)

\(\displaystyle = \ ln(|u|) - ln(|1+u|) + C\)

\(\displaystyle = \ ln\left (\dfrac{u}{1+u}\right ) + C\)

\(\displaystyle = \ ln\left (\dfrac{e^x}{1+e^x}\right ) + C\)

\(\displaystyle = \ ln\left (\dfrac{1}{e^{-x}+1}\right ) + C\)

\(\displaystyle = \ - ln(e^{-x}+1) + C\)

 

[Jason76]

My answer is the same as what your book gives.

Your working still has an integral in the final answer, and you have illegally put a function of x outside of the integral to balance a function put inside. You can only do this with constants.

Right, I know see what the problem is: The \(\displaystyle e^{x}\) is not a constant.

Of course, it's worth remembering that: \(\displaystyle e^{-x}(e ^{x}) = e^{0}\) and \(\displaystyle e^{0} = 1\) For instance, when the distributive property is done on the bottom, then the 1 turns into \(\displaystyle e^{-x}\) and the \(\displaystyle e^{x}\) becomes 1. Finally, the negative sign (constant) on the du dx doesn't balance out with the right side, so you have to balance it. You balance it with the reciprocal of -1 which is 1 over -1 which equals -1 or -.