1. tough integral?

Hey Kriz. Here's an integral for ya'.

$\displaystyle \int_{0}^{1}\frac{4x^{3}(1+x^{4(2006)})}{(1+x^{4}) ^{2008}}dx$

2. 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:

$\displaystyle = \frac{{\left( {1 - \frac{1} {{1 + x^4 }}} \right)^{2007} - \left( {\frac{1} {{1 + x^4 }}} \right)^{2007} }} {{2007}} = \left. {\frac{{x^{8028} - 1}} {{2007\left( {1 + x^4 } \right)^{2007} }}} \right|_0^1 = \frac{1} {{2007}}$

3. Originally Posted by galactus
Hey Kriz. Here's an integral for ya'.

$\displaystyle \int_{0}^{1}\frac{4x^{3}(1+x^{4(2006)})}{(1+x^{4}) ^{2008}}dx$
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 $\displaystyle \int_0^1 {\frac{{4x^3 \left( {1 + x^{4n} } \right)}} {{\left( {1 + x^4 } \right)^{n + 2} }}\,dx} = \frac{1} {{n + 1}}\quad\blacksquare$

4. Originally Posted by Krizalid
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 $\displaystyle \int_0^1 {\frac{{4x^3 \left( {1 + x^{4n} } \right)}} {{\left( {1 + x^4 } \right)^{n + 2} }}\,dx} = \frac{1} {{n + 1}}\quad\blacksquare$
that was... beautiful

5. 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 $\displaystyle \frac{1}{2007}$

6. It wasn't after all.

When I saw your problem, I think it was messy, but while solving it, it was getting interesting.

7. Yeah, I thought it was cool. I posted my way. It was kind of like yours, except you went one further and proive dthe general case. Show off

Now, it's time for me to

8. Originally Posted by Krizalid
Substitute $\displaystyle v = \frac{u} {{u + 1}}.$
Originally Posted by galactus

Now, sub in $\displaystyle t=1-\frac{1}{u}$
Wait, how did we "see" these substitutions?

9. 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.