3
$\begingroup$

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.

$\endgroup$
  • 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.