Question

find the following

$\int \: \frac{dx}{ {e}^{2x} - 3 {e}^{x} + 2 }$
​

Answer 1

Solution:

Given Integral:

$\displaystyle\rm=\int\dfrac{dx}{e^{2x}-3e^{x}+2}$

Let us assume that:

$\rm\longrightarrow u=e^{x}$

Therefore:

$\rm\longrightarrow du=e^{x}\:dx$

$\rm\longrightarrow dx=e^{-x}\:du$

So, the integral becomes:

$\displaystyle\rm=\int\dfrac{du}{u(u^{2}-3u+2)}$

$\displaystyle\rm=\int\dfrac{du}{u(u-1)(u-2)}$

Decompose the fraction into partial fraction. We get:

$\displaystyle\rm=\int\bigg[\dfrac{1}{2u}+\dfrac{-1}{u-1}+\dfrac{1}{2(u-2)}\bigg]\:du$

$\displaystyle\rm=\int\dfrac{1}{2u}\:du-\int\dfrac{1}{u-1}\:du+\int\dfrac{1}{2(u-2)}\:du$

$\displaystyle\rm=\dfrac{1}{2}\int\dfrac{1}{u}\:du-\int\dfrac{1}{u-1}\:du+\dfrac{1}{2}\int\dfrac{1}{(u-2)}\:du$

$\displaystyle\rm=\dfrac{1}{2}ln(|u|)-\int\dfrac{1}{u-1}\:du+\dfrac{1}{2}\int\dfrac{1}{(u-2)}\:du$

Now substitute v = u - 1.

$\rm\longrightarrow dv=dx$

Therefore, the integral becomes:

$\displaystyle\rm=\dfrac{1}{2}ln(|u|)-\int\dfrac{1}{v}\:dv+\dfrac{1}{2}\int\dfrac{1}{(u-2)}\:du$

$\displaystyle\rm=\dfrac{1}{2}ln(|u|)-ln(|v|)+\dfrac{1}{2}\int\dfrac{1}{(u-2)}\:du$

$\displaystyle\rm=\dfrac{1}{2}ln(|u|)-ln(|u-1|)+\dfrac{1}{2}\int\dfrac{1}{(u-2)}\:du$

Now substitute v = u - 2.

$\rm\longrightarrow dv=dx$

Therefore, the integral becomes:

$\displaystyle\rm=\dfrac{1}{2}ln(|u|)-ln(|u-1|)+\dfrac{1}{2}\int\dfrac{1}{v}\:dv$

$\displaystyle\rm=\dfrac{1}{2}ln(|u|)-ln(|u-1|)+\dfrac{1}{2}ln(|v|)+C$

$\displaystyle\rm=\dfrac{1}{2}ln(|u|)-ln(|u-1|)+\dfrac{1}{2}ln(|u-2|)+C$

Substitute back u = eˣ.

$\displaystyle\rm=\dfrac{1}{2}ln(e^{x})-ln(|e^{x}-1|)+\dfrac{1}{2}ln(|e^{x}-2|)+C$

$\displaystyle\rm=\dfrac{x}{2}-ln(|e^{x}-1|)+\dfrac{1}{2}ln(|e^{x}-2|)+C$

$\displaystyle\rm=\dfrac{\ln(|e^{x}-2|)+x}{2}-ln(|e^{x}-1|)+C$

Therefore:

$\displaystyle\rm\longrightarrow\int\dfrac{dx}{e^{2x}-3e^{x}+2}=\dfrac{\ln(|e^{x}-2|)+x}{2}-ln(|e^{x}-1|)+C$

Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$