improper integral

[ijd5000]

Start with the integral from 1 to infinity of (e^-x)/(e^(-x)+2)dx. I used substitution and got it to -ln(e^(-x) +2)+c . Next it somehow gets simplfied to x-ln(2e^x +1). I've tried everything i can think of.

 

[DrPhil]

Start with the integral from 1 to infinity of (e^-x)/(e^(-x)+2)dx.
I used substitution and got it to -ln(e^(-x) +2)+c . Yes, this looks good for the antiderivative
Next it somehow gets simplfied to x-ln(2e^x +1). I don't see where this came from.
I've tried everything i can think of. Did you try evaluating at upper and lower limits?

\(\displaystyle \displaystyle \int_1^\infty \dfrac{e^{-x}}{e^{-x}+2}dx\)

You didn't say what substitution you used - perhaps
......\(\displaystyle u = e^{-x}+2\), ... \(\displaystyle du = -e^{-x}\ dx \)
......\(\displaystyle x=1 \Rightarrow u=(1/e) +2\),....\(\displaystyle x=\infty \Rightarrow u=2\)
and the integral becomes

\(\displaystyle \displaystyle \int_{(1/e)+2}^2 \dfrac{-du}{u} = \int^{(1/e)+2}_2 \dfrac{du}{u} = \left. \ln u \right|^{(1/e)+2}_2 = \ln\left(\dfrac{1}{2e} + 1\right)\)

I get the same result if using your antiderivative between the limits x=1 and x=infinity.

 

[ijd5000]

\(\displaystyle \displaystyle \int_1^\infty \dfrac{e^{-x}}{e^{-x}+2}dx\)

You didn't say what substitution you used - perhaps
......\(\displaystyle u = e^{-x}+2\), ... \(\displaystyle du = -e^{-x}\ dx \)
......\(\displaystyle x=1 \Rightarrow u=(1/e) +2\),....\(\displaystyle x=\infty \Rightarrow u=2\)
and the integral becomes

\(\displaystyle \displaystyle \int_{(1/e)+2}^2 \dfrac{-du}{u} = \int^{(1/e)+2}_2 \dfrac{du}{u} = \left. \ln u \right|^{(1/e)+2}_2 = \ln\left(\dfrac{1}{2e} + 1\right)\)

I get the same result if using your antiderivative between the limits x=1 and x=infinity.

that is the substitution that i used. The answer to the problem is -1-ln(2)+ln(2e+1). I got the same answer as you after plugging in the limits, which is not one of the choices. It has to be equivalent but they used some weird simplification I've never seen.

 

[DrPhil]

that is the substitution that i used. The answer to the problem is -1-ln(2)+ln(2e+1). I got the same answer as you after plugging in the limits, which is not one of the choices. It has to be equivalent but they used some weird simplification I've never seen.

I don't see where the "-1" could come from. The evaluation has to give the difference of two logarithms. The difference of two logarithms may be expressed as the logarithm of a ratio.

Perhaps another of our tutors could comment?

 

[ijd5000]

wolfram says it too?

 

[daon2]

that is the substitution that i used. The answer to the problem is -1-ln(2)+ln(2e+1). I got the same answer as you after plugging in the limits, which is not one of the choices. It has to be equivalent but they used some weird simplification I've never seen.

\(\displaystyle \ln(2e+1)-1-\ln(2) = \ln(2e+1) -[\ln(e)+\ln(2)] = \ln(2e+1)-\ln(2e) = \ln\left(\dfrac{2e+1}{2e}\right) = \ln\left(\dfrac{1}{2e}+1\right)\)

 

[DrPhil]

wolfram says it too?

Lets try to manipulate that expression.

Note that x is the logarithm of e^x, and the difference of two logs is the log of the quotient.

\(\displaystyle \displaystyle x - \ln(2\ e^x + 1) = \ln e^x - \ln(2\ e^x + 1) = \ln \dfrac{e^x}{2\ e^x + 1} \)

Multiply both numerator and denominator by e^-x

..............\(\displaystyle = \ln \dfrac{1}{2 + e^{-x}} = -\ln(e^{-x}+2) = -\ln u\)

So the Wolfram answer for the indefinite integral IS the same as what we both got. But we have it in a form that is defined at x=infinity. And we have plugged in the limits several times, with the result

\(\displaystyle \displaystyle \ln \left(\dfrac{1}{2e} +1\right)\)

To make this agree with one of your multiple choices, you will have to do some algebra with the logs. We have simplified as far as it will go!

EDIT: If you would rather do arithmetic than algebra, note that our answer is a single number, 0.1688476...
You could evaluate the multiple choices to see which is exactly the same