- #1
Petar Mali
- 290
- 0
Homework Statement
Solve integral
[tex]\int \frac{(1+\Phi)+\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx[/tex]
where [tex]\Phi=const[/tex]
Homework Equations
The Attempt at a Solution
[tex]\int \frac{(1+\Phi)+\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx=\int\frac{1+\Phi}{(1+\Phi)-\Phi e^{-x}}dx+\int\frac{\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx[/tex]
[tex](1+\Phi)\int\frac{dx}{(1+\Phi)-\Phi e^{-x}}[/tex]
[tex](1+\Phi)-\Phi e^{-x}=t[/tex] [tex]1+\Phi-t=\Phi e^{-x}[/tex]
[tex]\Phi e^{-x}dx=dt[/tex]
[tex]\frac{1}{1+\Phi-t}=\frac{A}{1+\Phi-t}+\frac{B}{t}[/tex]
I got
[tex]A=B=\frac{1}{1+\Phi}[/tex]
So I got if I don't write constant
[tex](1+\Phi)\int\frac{dx}{(1+\Phi)-\Phi e^{-x}}=ln[\frac{(1+\Phi)-\Phi e^{-x}}{\Phi e^{-x}}][/tex]
For second integral I got without constant
[tex]\int\frac{\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx=ln[(1+\Phi)-\Phi e^{-x}][/tex]
So
[tex]\int \frac{(1+\Phi)+\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx=2ln[\frac{1+\Phi-\Phi e^{-x}}{\Phi e^{-x}}]+C[/tex]
Is this solution correct? Thanks for your answer!