Problem source

Show that
\[
\int_0^1 \frac{x^4}{1+x^2} \, \d x = \frac \pi {4} - \frac 23 \;.
\]
Determine the values of
\begin{questionparts}
\item $\displaystyle \int_0^1 x^3 \; \tan ^{-1} \left(\frac {1-x} {1+x} \right) \,\d x $
\item $\displaystyle \int_0^1 \frac {(1-y)^3} {(1+y)^5} \; {{\tan}^{-1} y}\, \d y$
\end{questionparts}

Solution source

\begin{align*}
&& \int_0^1 \frac{x^4}{1+x^2} \d x &= \int_0^1 \frac{(x^2-1)(1+x^2)+1}{x^2+1} \d x\\
&&&= \int_0^1 \frac{1}{1+x^2} \d x -\int_0^1 (1-x^2) \d x \\
&&&= \left [\tan^{-1}x \right]_0^1 - \left [x - \tfrac13x^3 \right]_0^1 \\
&&&= \frac{\pi}{4} - \frac23
\end{align*}

\begin{questionparts}
\item $\,$ \begin{align*}
&& I &= \int_0^1 x^3 \; \tan ^{-1} \left(\frac {1-x} {1+x} \right) \,\d x \\
&&&= \left [ \frac{x^4}{4}\tan ^{-1} \left(\frac {1-x} {1+x} \right) \right]_0^1 -\int_0^1 \frac{x^4}{4} \frac{1}{1 +\left(\frac {1-x} {1+x} \right) ^2 } \cdot \frac{-2}{(1+x)^2} \d x \\
&&&= \frac{1}{2} \int_0^1 \frac{x^4}{(1+x)^2+(1-x)^2} \d x \\
&&&= \frac{1}{4} \int_0^1 \frac{x^4}{1+x^2} \d x \\
&&&= \frac{\pi}{16} - \frac{1}{6}
\end{align*}

\item $\,$ \begin{align*}
&& J &=  \int_0^1 \frac {(1-y)^3} {(1+y)^5} \; {{\tan}^{-1} y}\, \d y \\
&&&= \left [ \frac {(y(1+y^2)} {(1+y)^4} \tan^{-1}y \right]_0^1 - \int_0^1 \frac {(y(1+y^2)} {(1+y)^4} \frac{1}{1+y^2} \d y \\
&&&= \frac{\pi}{32} - \int_0^1 \frac{y}{(1+y)^4} \d y \\
&&&= \frac{\pi}{32} - \left[ - \frac{3y+1}{6(1+y)^3} \right]_0^1 \\
&&&= \frac{\pi}{32} +\frac{4}{6 \cdot 8} - \frac{1}{6} \\
&&&= \frac{\pi}{32} - \frac{1}{12}
\end{align*}
\end{questionparts}