1. ## Integrating e^x

I am unsure on how to calculate the following 2 integrals

a) [1/(x^2)]*[e^(2/x)], with respect to xand
b) 1/(e^x + 1), with respect to x

Help with this will be greatly appreciated.

2. Originally Posted by Nas
I am unsure on how to calculate the following 2 integrals

a) [1/(x^2)]*[e^(2/x)], with respect to xand
b) 1/(e^x + 1), with respect to x

Help with this will be greatly appreciated.
Here you have

a) $\int{\left(\frac{1}{x^2}\right)e^{\frac{2}{x}}\,dx } = \int{e^{2x^{-1}}\,x^{-2}\,dx}$

$= -\frac{1}{2}\int{e^{2x^{-1}}(-2x^{-2})\,dx}$.

Now make the substitution $u = 2x^{-1}$ so that $du = -2x^{-2}\,dx$ and the integral becomes

$-\frac{1}{2}\int{e^u\,du}$

$= -\frac{1}{2}e^u + C$

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

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

3. Originally Posted by Nas
I am unsure on how to calculate the following 2 integrals

a) [1/(x^2)]*[e^(2/x)], with respect to xand
b) 1/(e^x + 1), with respect to x

Help with this will be greatly appreciated.
b) $\int{\frac{1}{e^x + 1}\,dx} = \int{\frac{e^x}{e^x(e^x + 1)}\,dx}$

$= \int{\left[\frac{1}{e^x(e^x + 1)}\right]e^x\,dx}$.

Now make the substitution $u = e^x + 1$ so that $e^x = u - 1$ and $du = e^x\,dx$.

The integral becomes

$\int{\frac{1}{(u - 1)u}\,du}$.

Now solve this using partial fractions.

4. would the answer for part b be x-log(e^x+1)?

1. Let $u = \dfrac{2}{x}$, then $\dfrac{du}{dx} = \dfrac{-2}{x^2} \Rightarrow {dx} = -\dfrac{x^2{du}}{2}$. Therefore $\int\dfrac{1}{x^2}e^{\frac{2}{x}}\;{dx} = -\int\left(\dfrac{e^u}{x^2}\right)\left(\dfrac{x^2} {2}\right)\;{du} = ...$
2. Let $u = e^x+1$, then $\dfrac{du}{dx} = e^x \Rightarrow {dx} = \dfrac{du}{e^x}.[/tex] Therefore [tex]\int\dfrac{1}{e^x+1}\;{dx} = \int\dfrac{1}{u(u-1)}\;{du} = ...$

5. Another solution for 2:

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

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

Let $u=e^{-x}+1$, to get :

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

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

6. Hello, Nas!

The General beat me to the solution . . . *sigh*

$(b)\;\;\int\frac{dx}{e^x+1}$

I have an equivalent answer: . $-\ln\left(1 + e^{-x}\right) + C$

And this answer can be simplified beyond all recognition . . .

$-\ln\left(1 + \frac{1}{e^x}\right) + C\;=\;-\ln\left(\frac{e^x+1}{e^x}\right) + C$

. . . . . . . . . . . . $=\;-\ln(e^x+1) + \ln(e^x) + C$

. . . . . . . . . . . . $=\;-\ln(e^x+1) + x\underbrace{\ln(e)}_{\text{This is 1}} + C$

. . . . . . . . . . . . $=\;x - \ln(e^x+1) + C$

7. Thanks for all the help.