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.
 A: You can use partial fractions: 
$$
\frac{1}{1-\theta(1-b)^2}=\frac12\,\left(\frac1{1-\sqrt{\theta}(1-b)}+\frac{1}{1+\sqrt{\theta}(1-b)}\right).
$$
Now, using the substitution $v=1-\sqrt{\theta}(1-b)$, we have $dv=\sqrt{\theta}\,db$, so the integral of the first half is 
\begin{align}
\int_0^1\frac{b^3}{1-\sqrt{\theta}(1-b)}\,db
&=\theta^{-1/2}\int_{1-\sqrt{\theta}}^1\frac{\left(\frac{v-1+\sqrt{\theta}}{\sqrt{\theta}} \right)^3}{v}\,dv
=\theta^{-2}\int_{1-\sqrt{\theta}}^1\frac{\left({v-1+\sqrt{\theta}} \right)^3}{v}\,dv\\ \ \\
&=\frac1{\theta^{2}}\,\int_{1-\sqrt{\theta}}^1\frac{\left(v^3-3v^2(1-\sqrt{\theta})+3v(1-\sqrt{\theta})^2+(1-\sqrt{\theta})^3 \right)}{v}\,dv\\ \ \\
&=\frac1{\theta^{2}}\,\int_{1-\sqrt{\theta}}^1{\left(v^2-3v(1-\sqrt{\theta})+3(1-\sqrt{\theta})^2+\frac{(1-\sqrt{\theta})^3}v \right)}{}\,dv\\ \ \\
&=\frac1{\theta^{2}}\,\left.\vphantom{\int}{\left(\frac{v^3}3-\frac{3v^2(1-\sqrt{\theta})}2+3v(1-\sqrt{\theta})^2+{(1-\sqrt{\theta})^3}\log v \right)}{}\right|_{1-\sqrt{\theta}}^1\\ \ \\
&=\frac1{\theta^{2}}\,{\left(\frac13-\frac{11(1-\sqrt{\theta})^3}6-\frac{3(1-\sqrt{\theta})}2+3(1-\sqrt{\theta})^2-{(1-\sqrt{\theta})^3}\log (1-\sqrt{\theta}) \right)}{}\\ \ \\
\end{align}
The second half can be calculated similarly. 
\begin{align}
\int_0^1\frac{b^3}{1+\sqrt{\theta}(1-b)}\,db
&=\theta^{-1/2}\int_1^{1+\sqrt{\theta}}\frac{\left(\frac{1-v+\sqrt{\theta}}{\sqrt{\theta}} \right)^3}{v}\,dv
=\theta^{-2}\int_1^{1+\sqrt{\theta}}\frac{\left({1-v+\sqrt{\theta}} \right)^3}{v}\,dv\\ \ \\
&=\frac1{\theta^{2}}\,\int_1^{1+\sqrt{\theta}}\frac{\left(-v^3+3v^2(1+\sqrt{\theta})-3v(1+\sqrt{\theta})^2+(1+\sqrt{\theta})^3 \right)}{v}\,dv\\ \ \\
&=\frac1{\theta^{2}}\,\int_1^{1+\sqrt{\theta}}{\left(-v^2+3v(1+\sqrt{\theta})-3(1+\sqrt{\theta})^2+\frac{(1+\sqrt{\theta})^3}v \right)}{}\,dv\\ \ \\
&=\frac1{\theta^{2}}\,\left.\vphantom{\int}{\left(-\frac{v^3}3+\frac{3v^2(1+\sqrt{\theta})}2-3v(1+\sqrt{\theta})^2+{(1+\sqrt{\theta})^3}\log v \right)}{}\right|_1^{1+\sqrt{\theta}}\\ \ \\
&=\frac1{\theta^{2}}\,{\left(-\frac13-\frac{7(1+\sqrt{\theta})^3}6-\frac{3(1+\sqrt{\theta})}2+3(1+\sqrt{\theta})^2+{(1+\sqrt{\theta})^3}\log (1+\sqrt{\theta}) \right)}{}.
\end{align}
Now we need to add the two integrals and divide by two. We get 
$$
\frac1{2\theta^2}\,\left(
\frac{2\theta^{3/2}}{3}-3\theta+2\sqrt\theta-{(1-\sqrt{\theta})^3}\log (1-\sqrt{\theta}) 
+
{(1+\sqrt{\theta})^3}\log (1+\sqrt{\theta})
\right).
$$
A: 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.
A: $\int_{0}^{1}\dfrac{b^3}{1-\theta(1-b)^2}db$. 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.
A: 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:
$\dfrac{-1}{\theta}\int_{0}^{1}(b+2+\dfrac{(3\theta+1)}{2\theta}\dfrac{2\theta-2\theta b}{1-\theta (1-b)^2}-\dfrac{(3\theta+1)}{2\theta}(\dfrac{4\theta -4\theta^2}{3\theta+1}+2\theta)\dfrac{1}{1-\theta (1-b)^2})db$
$=\dfrac{-1}{\theta}(\dfrac{1}{2}+2-\dfrac{(3\theta+1)}{2\theta}\ln|1-\theta|-\dfrac{\theta+3}{\sqrt \theta}(\ln|1+\sqrt \theta|-\ln|\sqrt {1-\theta}|))$
