I have been trying to solve the integral:
\begin{equation} \int \frac{((n-1)^2 - 4n \gamma e^{\frac{\gamma}{\lambda} t})^{\frac{1}{2}}}{\frac{1}{\gamma} + C e^{\frac{\gamma}{\lambda}t}} dt \end{equation}
where $\gamma , \lambda $ are positive constants.
So far, by using a substitution of variables \begin{equation} u = (n-1)^2 - 4n \gamma e^{\frac{\gamma}{\lambda} t} \end{equation}
I have got to the integral: \begin{equation} \int -4n \lambda \frac{u^{\frac{1}{2}}}{u^2 -2u (n^2 + 1) + (n^2 -1 )^2} du \end{equation}
but I am quite unsure as to how to proceed from here.
Any help or suggestions would be greatly appreciated, thank you!
I think the substitution u^(1/2) = v will work.
[p.s.] We can use partial fraction decomposition, which is the general way of integration of rational function.
$\int \frac{u^{1/2}}{u^2 - 2(n^2+1)u + (n^2-1)^2} du = \int \frac{2v^2}{v^4 - 2(n^2+1)v^2 + (n^2-1)^2} dv = \int \frac{2v^2}{(v+1-n)(v+1+n)(v-1-n)(v-1+n)} dv = \int ( \frac{n+1}{4n}\frac{1}{v-n-1} -\frac{n+1}{4n}\frac{1}{v+n+1} +\frac{n-1}{4n}\frac{1}{v+n-1} -\frac{n-1}{4n}\frac{1}{v-n+1} ) dv$