<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] Because of the $b^3$ in numerator, I should rather start with
$$\frac{b^{3}}{1 - \theta(1-b)^{2}}=\frac{b+2}{\theta }+\frac{b (3 \theta +1)-2 \theta +2}{\theta  \left(\theta(b-1)^2 -1\right)}$$
$$\frac{b^{3}}{1 - \theta(1-b)^{2}}=\frac{b+2}{\theta }+\frac{b (3 \theta +1)-2 \theta +2}{\theta  \left(\theta( b-1)^2 -1\right)}=\frac{b+2}{\theta }+\dfrac{\left(3\theta+1\right)\left(b-1\right)}{\theta\left(b-1\right)^2-1}+\dfrac{\theta+3}{\theta\left(b-1\right)^2-1}$$
Now, we face quite simple antiderivatives at the price of very simple substitutions.
</s>