Hey Kriz. Here's an integral for ya'.
$\displaystyle \int_{0}^{1}\frac{4x^{3}(1+x^{4(2006)})}{(1+x^{4}) ^{2008}}dx$
for this tough integral we'll use the following substitution:
$\displaystyle \begin{gathered}
t = \frac{1}
{{1 + x^4 }} \hfill \\
\hfill \\
dt = \frac{{ - 4x^3 }}
{{\left( {1 + x^4 } \right)^2 }}dx \hfill \\
\end{gathered} $
$\displaystyle
\int\limits_0^1 { - t^{2006} \left[ {1 + \left( {\frac{{1 - t}}
{t}} \right)^{2006} } \right]} dt = \int\limits_0^1 { - t^{2006} - \left( {1 - t} \right)^{2006} } dt = - \frac{{t^{2007} }}
{{2007}} + \frac{{\left( {1 - t} \right)^{2007} }}
{{2007}}
$
now back substitute:
$\displaystyle
= \frac{{\left( {1 - \frac{1}
{{1 + x^4 }}} \right)^{2007} - \left( {\frac{1}
{{1 + x^4 }}} \right)^{2007} }}
{{2007}} = \left. {\frac{{x^{8028} - 1}}
{{2007\left( {1 + x^4 } \right)^{2007} }}} \right|_0^1 = \frac{1}
{{2007}}
$
More generally: let $\displaystyle n\ge0,$ so $\displaystyle \int_0^1 {\frac{{4x^3 \left( {1 + x^{4n} } \right)}}
{{\left( {1 + x^4 } \right)^{n + 2} }}\,dx} = \frac{1}
{{n + 1}}.$
First make the obvious substitution: $\displaystyle u=x^4,$ so
$\displaystyle \int_0^1 {\frac{{4x^3 \left( {1 + x^{4n} } \right)}}
{{\left( {1 + x^4 } \right)^{n + 2} }}\,dx} = \int_0^1 {\frac{{1 + u^n }}
{{(1 + u)^{n + 2} }}\,du} .$
Splittin' the original ratio into two fractions we get an easy integral whose value is $\displaystyle \frac{{1 - 2^{ - n - 1} }}
{{n + 1}}.$ Now here's the interesting part: the second piece.
We have $\displaystyle \int_0^1 {\frac{{u^n }}
{{(1 + u)^{n + 2} }}\,du} .$ Substitute $\displaystyle v = \frac{u}
{{u + 1}}.$ After some simple calculations our integral becomes to
$\displaystyle \int_0^{1/2} {v^n \,dv} = \left. {\frac{{v^{n + 1} }}
{{n + 1}}} \right|_0^{1/2} = \frac{{2^{ - n - 1} }}
{{n + 1}}.$
Finally, put these things together and we have $\displaystyle \int_0^1 {\frac{{4x^3 \left( {1 + x^{4n} } \right)}}
{{\left( {1 + x^4 } \right)^{n + 2} }}\,dx} = \frac{1}
{{n + 1}}\quad\blacksquare$
Very nice and elegant. Wasn't tough afterall, was it?.
Here's what I done:
Broke it up
$\displaystyle \int_{0}^{1}\frac{4x^{3}}{(1+x^{4})^{2008}}dx+\int _{0}^{1}\frac{4x^{3}\cdot{x^{4(2006)}}}{(1+x^{4})^ {2008}}dx$
sub $\displaystyle u=1+x^{4}$ in first part.
$\displaystyle \int_{1}^{2}\frac{1}{u^{2008}}du$
Second part beomes:
$\displaystyle \int_{1}^{2}\frac{(u-1)^{2006}}{u^{2008}}du=\int_{1}^{2}\frac{(1-\frac{1}{u})^{2006}}{u^{2}}du$
Now, sub in $\displaystyle t=1-\frac{1}{u}$
and the last one becomes:
$\displaystyle \int_{0}^{\frac{1}{2}}t^{2006}dt$
Put it all together and you do indeed get $\displaystyle \frac{1}{2007}$
When I got $\displaystyle \int_{0}^{1}{\frac{u^{n}}{(1+u)^{n+2}}\,du},$ I made $\displaystyle \frac{u^{n}}{(1+u)^{n+2}}=\left( \frac{u}{1+u} \right)^{n}\cdot \frac{1}{(1+u)^{2}}.$ Since by differentiating $\displaystyle \frac{u}{1+u}=\frac{1+u-1}{1+u}=1-\frac{1}{1+u}$ it immediately yields a part of the integrand and from there you set the given substitution.