The discussion centers on the integration of the expression $\int \frac{e^x - e^{-x}}{e^{-x} + 1} dx$. A user initially attempts to simplify the integral by multiplying by $e^x$, but receives guidance to factor $e^x$ out of the numerator instead. The recommended substitution involves letting $u = e^{-x}$, which transforms the integral into $-\int \frac{1 - u^2}{u + 1} du$, facilitating easier integration. This approach emphasizes the importance of strategic manipulation in integral calculus.
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markosheehan said: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:markosheehan said: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
HallsofIvy said: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 $1- u^2= (1- u)(1+ u)$.