<s>[INST] 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.
 [/INST] 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*})
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