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.