Evaluate \[ \int \frac{1+e^x}{1-e^x}\,dx. \]
Solution
Put \(u=e^x\), so \(dx=du/u\). Then \[ \frac{1+u}{u(1-u)}=\frac1u+\frac2{1-u}. \] Hence \[ \int \frac{1+e^x}{1-e^x}\,dx =\ln u-2\ln|1-u|+C =\boxed{x-2\ln|1-e^x|+C}. \]
Antiderivative of $\frac{1+e^x}{1-e^x}$
P-TO7UO
- Area
- Prelim
- Topics
- Integrals, u-Substitution
- Status
- draft