The following question is only related to eq.1 below and rest of the eqs. are just there to let know why I want, what I want
$$\mathcal{L_I} = \frac{1}{1 + xT(r/n)^\alpha P_o/P_c} \ \ \ \ \ \ \ \ \ \ (1)$$
$$ f_R(r) = 2\pi\lambda re^{-\pi r^2}\ t \ \ \ \ \ \ \ \ \ \ \ \ (2)$$
$$P = \int^\infty_0\mathcal{L_I}f_R(r) dr \ \ \ \ \ \ \ \ \ (3)$$
Now, I want to simplify (1) so that $r$-term comes in the numerator and no term in the denominator of eq. 1 contains $r$. This is just concerning the way I want to put the equation, can someone
\begin{align} \mathcal{L_I} &= \frac{1}{1 + xT(r/n)^\alpha P_o/P_c}\\ \frac{1}{\mathcal{L_I}} &= 1 + xT(r/n)^\alpha P_o/P_c\\ \frac{1}{\mathcal{L_I}} - 1 &= xT(r/n)^\alpha P_o/P_c\\ \frac{1}{x\mathcal{L_I}} - \frac{1}{x} &= xT(r/n)^\alpha P_o/P_c\\ \frac{1}{xT\mathcal{L_I}} - \frac{1}{xT} &= (r/n)^\alpha P_o/P_c\\ \frac{P_c}{P-o xT\mathcal{L_I}} - \frac{P_c}{P-o xT} &= (r/n)^\alpha \\ \left( \frac{P_c}{P-o xT\mathcal{L_I}} - \frac{P_c}{P-o xT} \right)^\frac{1}{\alpha} &= r/n\\ n \cdot \left( \frac{P_c}{P-o xT\mathcal{L_I}} - \frac{P_c}{P-o xT} \right)^\frac{1}{\alpha} &= r\\ \end{align}