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.

$\int_{0}^{1}\dfrac{b^3}{1-\theta(1-b)^2}db$. Integrating by parts, taking u=b and dv as the integral of the rest we have:
$(b\int\dfrac{b^2}{1-\theta(1-b)^2}db)\Big|_0^1  -\int_{0}^{1}(\int\dfrac{b^2}{1-\theta(1-b)^2}db)db$.
Lets solve the first integral.
$\int\dfrac{b^2}{1-\theta(1-b)^2}db=\dfrac{-1}{\theta}\int\dfrac{1-\theta+2\theta b-\theta b^2-1+\theta-2\theta b}{1-\theta(1-b)^2}db$
$=\dfrac{-1}{\theta}\int(1+\dfrac{-1+\theta -2\theta b}{1-\theta(1-b)^2})db$
$=\dfrac{-1}{\theta}\int(1+\dfrac{2\theta(1 - b)}{1-\theta(1-b)^2}+\dfrac{-1-\theta}{1-\theta(1-b)^2})db$
Taking $\sqrt\theta(1-b)=sen(t)$ in the last term
$=\dfrac{-1}{\theta}(\int(1+\dfrac{2\theta(1 - b)}{1-\theta(1-b)^2})db -\dfrac{1}{\sqrt{\theta}}(-1-\theta)\int \frac{cos(t)}{1-sen^2(t)}dt)$
We get $-\dfrac{1}{\theta}(b+\ln|1-\theta(1-b)|+\dfrac{(1+\theta)}{\sqrt{\theta}}(\ln|1+\sqrt \theta(1-b)|-\ln|\sqrt{1-\theta(1-b)^2}|)$
For the second integral we have to integrate the last result, and are only logarithms.
