MHB Integrating e^X - e^-x/e^-x+1 dx

[markosheehan]

how do you integrate e^X-e^-x/e^-x+1 dx
i am trying multiplying by e^x and trying to make it into no fraction but i am having no luck

 

[kaliprasad]

how do you integrate e^X-e^-x/e^-x+1 dx
i am trying multiplying by e^x and trying to make it into no fraction but i am having no luck

do you mean integrate

$\dfrac{e^x - e^{-x}}{e^{-x}+1}$ or $e^x - \dfrac{e^{-x}}{e^{-x}+1}$

 

[markosheehan]

The first one

 

[HallsofIvy]

how do you integrate e^X-e^-x/e^-x+1 dx
i am trying multiplying by e^x and trying to make it into no fraction but i am having no luck

Instead of multiplying by e^x, I would factor an e^x out of the numerator:
$\int \frac{e^x- e^{-x}}{e^{-x}+ 1}dx= \frac{1- e^{-2x}}{e^{-x}+ 1} e^xdx$

Let $u= e^{-x}$ so $du= -e^{-x}dx$. The integral becomes $-\int\frac{1- u^2}{u+ 1}du$. I presume you know that $1- u^2= (1- u)(1+ u)$.

 

[Prove It]

Instead of multiplying by e^x, I would factor an e^x out of the numerator:
$\int \frac{e^x- e^{-x}}{e^{-x}+ 1}dx= \frac{1- e^{-2x}}{e^{-x}+ 1} e^xdx$

Let $u= e^{-x}$ so $du= -e^{-x}dx$. The integral becomes $-\int\frac{1- u^2}{u+ 1}du$. I presume you know that $1- u^2= (1- u)(1+ u)$.

Why would you factor out e^x to be used in your differential when you need e^(-x)?