Solving Integral with Constant \Phi

[Petar Mali]

Homework Statement

Solve integral

\int \frac{(1+\Phi)+\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx

where \Phi=const

Homework Equations

The Attempt at a Solution

\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

(1+\Phi)\int\frac{dx}{(1+\Phi)-\Phi e^{-x}}

(1+\Phi)-\Phi e^{-x}=t 1+\Phi-t=\Phi e^{-x}

\Phi e^{-x}dx=dt

\frac{1}{1+\Phi-t}=\frac{A}{1+\Phi-t}+\frac{B}{t}

I got

A=B=\frac{1}{1+\Phi}

So I got if I don't write constant

(1+\Phi)\int\frac{dx}{(1+\Phi)-\Phi e^{-x}}=ln[\frac{(1+\Phi)-\Phi e^{-x}}{\Phi e^{-x}}]

For second integral I got without constant

\int\frac{\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx=ln[(1+\Phi)-\Phi e^{-x}]


So


\int \frac{(1+\Phi)+\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx=2ln[\frac{1+\Phi-\Phi e^{-x}}{\Phi e^{-x}}]+C

Is this solution correct? Thanks for your answer!

 

[rl.bhat]

Rewrite the given problem as
1 + [2φ*e^-x/(1 + φ - φe^-x)...(1)
If t = 1 + φ - φe^-x
dt = ...?
Substitute these values in eq. 1 and find integration.