<s>[INST] How do we integrate the expression $\frac{b^{3}}{1 - \theta(1-b)^{2}}$ where $0\leq b\leq 1$ and $\theta\in(0,1)$? MY ATTEMPT
In order to solve this integral, I have tried using the integration by parts method, as suggested by the expression
\begin{align}
\int_{0}^{1}\frac{b^{3}}{1-\theta(1-b)^{2}}\mathrm{d}b = \int_{0}^{1}\frac{b^{2}}{1-\theta(1-b)^{2}}\mathrm{d}b - \int_{0}^{1}\left[\int\frac{b^{2}}{1-\theta(1-b)^{2}}\mathrm{d}b\right]\mathrm{d}b
\end{align}
Based on it, the first summand is given by
\begin{align}\label{eq8}
\int_{0}^{1}\frac{b^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b & = \int_{0}^{1}\frac{(1 - 2b + b^{2}) - (1 - 2b)}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = \int_{0}^{1}\frac{(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b - \int_{0}^{1}\frac{(2-2b) - 1}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = \int_{0}^{1}\frac{(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b - 2\int_{0}^{1}\frac{1-b}{1 - \theta(1-b)^{2}}\mathrm{d}b + \int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b
\end{align}
The first of the last three integrals is given by
\begin{align}
\int_{0}^{1}\frac{(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b & = \frac{1}{\theta}\int_{0}^{1}\frac{\theta(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = -\frac{1}{\theta}\int_{0}^{1}\frac{1 - \theta(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b + \frac{1}{\theta}\int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = -\frac{1}{\theta} + \frac{1}{\theta}\int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b
\end{align}
According to the substitution $u = \sqrt{\theta}(1-b)$, it results that
\begin{align}\label{eq10}
\int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b = \frac{1}{\sqrt{\theta}}\int_{0}^{\sqrt{\theta}}\frac{1}{1 - u^{2}}\mathrm{d}u = \frac{1}{2\sqrt{\theta}}\ln\left|\frac{1+\sqrt{\theta}}{1-\sqrt{\theta}}\right|
\end{align}
Finally, based on the same substitution, we have
\begin{align}\label{eq11}
\int_{0}^{1}\frac{1-b}{1 - \theta(1-b)^{2}}\mathrm{d}b = \frac{1}{\theta}\int_{0}^{\sqrt{\theta}}\frac{u}{1-u^{2}}\mathrm{d}u = -\frac{1}{\theta}\ln|1-\theta^{2}|
\end{align}
Combining all these results, we get the first summand from the integration by parts method. The problem arises when I try to determine the second part.
 [/INST] In a similar way, we can make this:
$b^3=\dfrac{-1}{\theta}(b-\theta b+2\theta b^2-\theta b^3-b+\theta b-2\theta b^2)$
$\dfrac{b^3}{1-\theta (1-b)^2}=\dfrac{-1}{\theta}(b+\dfrac{-b+\theta b-2\theta b^2}{1-\theta (1-b)^2})$
We can write the last term as $2\dfrac{\frac{-b}{2}+\frac{b\theta}{2}-\theta b^2}{1-\theta (1-b)^2}=2 \dfrac{1-\theta +2\theta b-\theta b^2-1+\theta-2\theta b -\frac{b}{2}+\frac{b\theta}{2}}{1-\theta (1-b)^2}$
$=2(1+\dfrac{-1+\theta-2\theta b -\frac{b}{2}+\frac{b\theta}{2}}{1-\theta (1-b)^2})$
For the last numerator follows $-1+\theta+\dfrac{-(3\theta+1)}{2}b=\dfrac{1}{2\theta}(-2\theta+2\theta^2+\dfrac{-(3\theta+1)}{2}2\theta b)$
$=\dfrac{(3\theta+1)}{4\theta}(\dfrac{-4\theta +4\theta^2}{3\theta+1}-2\theta b+2\theta -2\theta)$
So, we have, finally:
$\dfrac{-1}{\theta}\int_{0}^{1}(b+2+\dfrac{(3\theta+1)}{2\theta}\dfrac{2\theta-2\theta b}{1-\theta (1-b)^2}-\dfrac{(3\theta+1)}{2\theta}(\dfrac{4\theta -4\theta^2}{3\theta+1}+2\theta)\dfrac{1}{1-\theta (1-b)^2})db$
$=\dfrac{-1}{\theta}(\dfrac{1}{2}+2-\dfrac{(3\theta+1)}{2\theta}\ln|1-\theta|-\dfrac{\theta+3}{\sqrt \theta}(\ln|1+\sqrt \theta|-\ln|\sqrt {1-\theta}|))$
</s>