интеграл 2

[Гост]

[tex]\int\frac{e^{x}+1}{e^{x}-1}dx[/tex]=?

[Sup3rlum]

Ето два начина:

1.

$\int \frac{e^x+1}{e^x-1}=dx \Rightarrow e^x-1=t \Rightarrow e^xdx=dt \Rightarrow dx = \frac{dt}{t+1}$

$\int \frac{t+2}{t}.\frac{dt}{t+1}$ с разделяне на дробите:

$\int \frac{2}{t}-\frac{1}{t+1}dt=2ln(t)-ln(t+1)+C$

$2ln(e^x-1)-x+C$


2.

$\int \frac{e^x+1}{e^x-1}dx=\int \frac{e^x-1+2}{e^x-1}dx=\int 1 + \frac{2}{e^x-1}dx \Rightarrow e^x-1=t \Rightarrow e^xdx=dt \Rightarrow dx=\frac{dt}{t+1}$
$x+\int \frac{2}{t}.\frac{dt}{t+1}=x+\int \frac{2}{t}-\frac{2}{t+1}dt$
$x+2ln(t)-2ln(t+1)+C=2ln(t)-x+C$