Home › Forums › Math questions › Calculus questions

Integrate

Posted January 21st, 2010 by alpha
in

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

‹ Integrate Integrate ›

Comments

answer

On January 21st, 2010 Structure says:

$ \int \frac{e^x-1}{e^x+1}dx = \int \frac{e^x-1}{e^x(e^x+1)}e^xdx==2\ln|e^x+1|-\ln|e^x|+\Mathcal{C}==\int\frac{y-1}{y(y+1)}dy=\int\frac{2y-y-1}{y(y+1)}=\int(\frac{2}{y+1}-\frac{1}{y})dy=2\ln|y+1|-\ln|y|+\Mathcal{C} $

The answer is ln(e^x+1)+ln((1+e^(-x))+C

On January 27th, 2010 alpha says:

Your answer is the same!!

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

-----------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

Back to top