Solve $\int\limits_0^{1/\sqrt{2}} \frac{au^2}{5(1-u^2)^2}du = 1$ for $a$ Problem:
The function $f_U(u) = \frac{au^2}{5(1-u^2)^2}$ is a probability density for the random variable $U$, which is non-zero on the interval $(0, \frac{1}{\sqrt{2}})$. I am supposed to find the value $a$.
I understand that that amounts to solving the equation $\int_0^{\frac{1}{\sqrt{2}}} \frac{au^2}{5(1-u^2)^2}du = 1$ for $a$.
Hint:
The hint I am given is to use integration by parts on the left hand side of $$\int_0^{\frac{1}{\sqrt{2}}} \frac{1}{1-u^2}du = \log(1 + \sqrt2)$$
My Attempt:
I don't understand how to use this hint. I know how to integrate by parts.
If the standard expression for integrating by parts is $st\bigg\rvert_0^{\frac{1}{\sqrt{2}}} - \int_0^{\frac{1}{\sqrt{2}}}tds$, then I guess I am supposed to find the appropriate $s$ and $dt$ in the expression $\dfrac{u^2}{(1-u^2)^2}$? In particular, I am supposed to choose $s$ and $dt$ such that $tds = \dfrac{1}{1-u^2}$?
I am led to believe that this is not a particularly nasty integral, though the online integral calculators do some sort of partial fraction decomposition that looks like a mess.
Another idea I had was to do a trigonometric substitution, but I don't think that will work.
 A: HINT:
Let $u=\sin\theta\implies du=\cos\theta \ d \theta$ $$\int_0^{\frac{1}{\sqrt{2}}} \frac{au^2}{5(1-u^2)^2}du =\int_0^{\pi/4}\frac{a\sin^2\theta }{5\cos^4\theta}\cos\theta d\theta$$
$$=\frac a5\int_0^{\pi/4}\frac{(1-\cos^2\theta) }{\cos^3\theta} d\theta$$
$$=\frac a5\int_0^{\pi/4}(\sec^3\theta-\sec\theta) \ d\theta$$
$$=\frac a5\left(\frac12\sec\theta\tan\theta-\frac12\ln|\sec\theta+\tan\theta|\right)_0^{\pi/4} $$
$$=\frac{a}{10}\left(\sqrt{2}-\ln(\sqrt{2}+1)\right)$$
A: Note
\begin{align}
\int_0^{1/\sqrt{2}} \frac{u^2du}{5(1-u^2)^2}
& =\frac1{10} \int\limits_0^{1/\sqrt{2}} u\> d\left(\frac{1}{1-u^2}\right)\\
&=\frac{u}{10(1-u^2)}\bigg|_0^{1/\sqrt2}-\frac 1{10} \int\limits_0^{1/\sqrt{2}} \frac{du}{1-u^2}\\
&= \frac{1}{5\sqrt2}-\frac 1{10}\ln(1+\sqrt2)
\end{align}
A: Hint (using your method and without the limits to keep it simple)
Let $u=\frac{1}{1-x^2}$ and $dv=dx$, then using integration by parts $\left(\int u \, dv=uv -\int v \, du\right)$, we get
\begin{align*}
\int\frac{1}{1-x^2} \, dx & =\frac{x}{1-x^2}-\int x \frac{(-2x)}{(1-x^2)^2} \, dx\\
& =\frac{x}{1-x^2}+2\int  \frac{x^2}{(1-x^2)^2} \, dx
\end{align*}
So we have
$$\color{red}{\int  \frac{x^2}{(1-x^2)^2} \, dx}=\frac{1}{2}\left[\color{blue}{\int\frac{1}{1-x^2}\, dx}-\frac{x}{1-x^2}\right].$$
The integral in red is the one you want and you already have the value for the integral in blue.
The integral on the right hand side can be done with partial fractions
$$\color{blue}{\int \frac{1}{1-x^2} \, dx}=\frac{1}{2}\left[\int \frac{1}{1-x} \, dx + \int \frac{1}{1+x} \, dx\right]=\frac{1}{2}\ln \left|\frac{1+x}{1-x}\right|+c.$$
I just added the last steps to show how it can be done.
A: Start by pulling the $\frac{a}{5}$ out of the integral.  You know how to factor the denominator into linear factors, so perform partial fraction decomposition.
$$  \frac{u^2}{(1-u^2)^2} = \frac{1/4}{u-1} + \frac{1/4}{(u-1)^2} + \frac{-1/4}{u+1} + \frac{1/4}{(u+1)^2}  \text{.}  $$
The antiderivatives of these terms are, respectively,

*

*$\frac{1}{4} \ln|u-1| + C$,

*$\frac{1}{4} \frac{-1}{u-1} + C$,

*$\frac{-1}{4} \ln|u+1| + C$, and

*$\frac{1}{4} \frac{-1}{u+1} + C$.

Then your integral is
$$  \frac{a}{20}\left( \left( \ln\left|\frac{1}{\sqrt{2}} - 1\right| + \frac{-1}{\frac{1}{\sqrt{2}} - 1} - \ln\left|\frac{1}{\sqrt{2}} + 1\right| + \frac{-1}{\frac{1}{\sqrt{2}} + 1} \right) - \left( \ln\left|0 - 1\right| + \frac{-1}{0 - 1} - \ln\left|0 + 1\right| + \frac{-1}{0 + 1} \right) \right)  $$
$$  = \frac{a}{20} \left( 2\sqrt{2} +\ln \frac{1-1/\sqrt{2}}{1+1/\sqrt{2}} \right)  $$
$$  = \frac{a}{10} \left( \sqrt{2} +\ln (\sqrt{2}-1) \right)  $$
(Using an identity and the fact that the (inverse) hyperbolic tangent is an odd function, we can see this is equivalent to other answers here.)
(For the last bit:  \begin{align*}
\ln \frac{1-1/\sqrt{2}}{1+1/\sqrt{2}}
    &= 2 \ln \sqrt{\frac{1-1/\sqrt{2}}{1+1/\sqrt{2}} }  \\
    &= 2 \ln \sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1} }  \\
    &= 2 \ln \sqrt{\frac{(\sqrt{2}-1)^2}{(\sqrt{2}+1)(\sqrt{2}-1)} }  \\
    &= 2 \ln \sqrt{\frac{(\sqrt{2}-1)^2}{2-1} }  \\
    &= 2 \ln \left| \sqrt{2}-1 \right| \\
    &= 2 \ln ( \sqrt{2}-1 )  \text{.}
\end{align*})
A: $$\int\limits_{u=0}^{1/\sqrt{2}} \frac{a u^2}{5 (1-u^2)^2}\ du = \frac{1}{10} a \left(\sqrt{2}-\tanh ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right)$$
Then solve
$$\frac{1}{10} a \left(\sqrt{2}-\tanh ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right) = 1$$
to find:
$$a = \frac{10}{\sqrt{2}-\tanh ^{-1}\left(\frac{1}{\sqrt{2}}\right)}$$
