I am trying to solve what looked like a simple integral but I got a bit stuck. The integral is : \begin{equation} \int_0^x \frac{ab(1-e^{-ct})}{d-\frac{b(1-e^{-ct})}{c}}dt \end{equation}

I tried change the change of variable $y=1-e^{-ct}$ and got :

\begin{equation} \int_0^{1-e^{-ct}} \frac{1}{y+\frac{cd}{by}-\frac{cd}{b}-1}dt \end{equation}

Now I am looking into changing again the variable $z=y+\frac{cd}{by}-\frac{cd}{b}-1$ but I am troubled by the lower limit of integration (when y=0). Can someone give this a try maybe you spot a better way of solving the thing. Thanks.

  • 3
    $\begingroup$ With your first change of variable I got $ab\int\limits_{0}^{1-{{e}^{-cx}}}{\frac{ydy}{(cd-by)(1-y)}}$. Now do partial fractions and integrate. $\endgroup$ – Paul Sep 24 '14 at 15:14
  • $\begingroup$ yes, I made a stupid mistake. Thanks Paul. $\endgroup$ – KAT Sep 24 '14 at 17:30

After your first change of variable you get

$$ac \int \dfrac{y}{\frac{cd}{b}-y} \dfrac{1}{(1-y)}dy$$

Now forgetting about the constants for a moment,

$$\int \dfrac{y}{(A-y)(B-y)} dy = \dfrac{1}{A-B}\int \dfrac{B}{B-y} - \dfrac{A}{A-y} dy$$

Then go back and substitute.


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