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) $\displaystyle \int{\left(\frac{1}{x^2}\right)e^{\frac{2}{x}}\,dx } = \int{e^{2x^{-1}}\,x^{-2}\,dx}$
$\displaystyle = -\frac{1}{2}\int{e^{2x^{-1}}(-2x^{-2})\,dx}$.
Now make the substitution $\displaystyle u = 2x^{-1}$ so that $\displaystyle du = -2x^{-2}\,dx$ and the integral becomes
$\displaystyle -\frac{1}{2}\int{e^u\,du}$
$\displaystyle = -\frac{1}{2}e^u + C$
$\displaystyle = -\frac{1}{2}e^{2x^{-1}} + C$
$\displaystyle =-\frac{1}{2}e^{\frac{2}{x}} + C$.
b) $\displaystyle \int{\frac{1}{e^x + 1}\,dx} = \int{\frac{e^x}{e^x(e^x + 1)}\,dx}$
$\displaystyle = \int{\left[\frac{1}{e^x(e^x + 1)}\right]e^x\,dx}$.
Now make the substitution $\displaystyle u = e^x + 1$ so that $\displaystyle e^x = u - 1$ and $\displaystyle du = e^x\,dx$.
The integral becomes
$\displaystyle \int{\frac{1}{(u - 1)u}\,du}$.
Now solve this using partial fractions.
- Let $\displaystyle u = \dfrac{2}{x}$, then $\displaystyle \dfrac{du}{dx} = \dfrac{-2}{x^2} \Rightarrow {dx} = -\dfrac{x^2{du}}{2}$. Therefore $\displaystyle \int\dfrac{1}{x^2}e^{\frac{2}{x}}\;{dx} = -\int\left(\dfrac{e^u}{x^2}\right)\left(\dfrac{x^2} {2}\right)\;{du} = ...$
- Let $\displaystyle u = e^x+1$, then $\displaystyle \dfrac{du}{dx} = e^x \Rightarrow {dx} = \dfrac{du}{e^x}.[/Math] Therefore [Math]\int\dfrac{1}{e^x+1}\;{dx} = \int\dfrac{1}{u(u-1)}\;{du} = ... $
Another solution for 2:
$\displaystyle \int \frac{dx}{e^x+1} = \int \frac{1}{e^x+1} \, \frac{e^{-x}}{e^{-x}} \, dx$
$\displaystyle =\int \frac{e^{-x}}{e^{-x}+1} \, dx$
Let $\displaystyle u=e^{-x}+1$, to get :
$\displaystyle - \int \frac{du}{u} = -ln|u|+C = -ln(e^{-x}+1) + C$
$\displaystyle =ln \left( \frac{1}{e^{-x}+1} \right) + C$
Hello, Nas!
The General beat me to the solution . . . *sigh*
$\displaystyle (b)\;\;\int\frac{dx}{e^x+1} $
I have an equivalent answer: .$\displaystyle -\ln\left(1 + e^{-x}\right) + C$
And this answer can be simplified beyond all recognition . . .
$\displaystyle -\ln\left(1 + \frac{1}{e^x}\right) + C\;=\;-\ln\left(\frac{e^x+1}{e^x}\right) + C$
. . . . . . . . . . . . $\displaystyle =\;-\ln(e^x+1) + \ln(e^x) + C$
. . . . . . . . . . . . $\displaystyle =\;-\ln(e^x+1) + x\underbrace{\ln(e)}_{\text{This is 1}} + C$
. . . . . . . . . . . . $\displaystyle =\;x - \ln(e^x+1) + C$