This is another question that I'm having trouble solving... is it also bad?
$\displaystyle
\int{\frac{x^2}{x^4+1}}dx
$
Thanks.
I really should find something better to do with my time. Well, at least the LaTeX bashing was fun and it looks vaguely cool.
I actually believe that by some really farfetched coincidence my answer is right.
The denominator can factor as:
$\displaystyle \int \frac{x^2}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}\,dx$
You can then use partial fractions lol
$\displaystyle =\int \left(\frac{\sqrt{2}x}{4(x^2-\sqrt{2}x+1)}-\frac{\sqrt{2}x}{4(x^2+\sqrt{2}x+1)}\right)\,dx$
$\displaystyle =\frac{\sqrt{2}}{4}\int \frac{x}{x^2-\sqrt{2}x+1}\,dx-\frac{\sqrt{2}}{4}\int \frac{x}{x^2+\sqrt{2}x+1}\,dx$
$\displaystyle =\frac{\sqrt{2}}{4}\int \frac{x}{(x-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}\,dx-\frac{\sqrt{2}}{4}\int \frac{x}{(x+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}\,dx$
Using $\displaystyle \int \frac{z}{a^2+z^2}\,dz=\frac{1}{2}\ln(a^2+z^2)+C$ and $\displaystyle \int \frac{\,dz}{a^2+z^2} = \frac{1}{a}\tan^{-1}{\frac{z}{a}}+C$, let $\displaystyle u = x+\frac{\sqrt{2}}{2}$ and $\displaystyle v = x-\frac{\sqrt{2}}{2}$, then the integral becomes
$\displaystyle \frac{\sqrt{2}}{4}\int \frac{u-\frac{\sqrt{2}}{2}}{u^2+\frac{1}{2}}\,du-\frac{\sqrt{2}}{4}\int \frac{v+\frac{\sqrt{2}}{2}}{v^2+\frac{1}{2}}\,dv$
$\displaystyle =\left(\frac{\sqrt{2}}{4}\int \frac{u}{u^2+\frac{1}{2}}\,du - \frac{\sqrt{2}}{4}\int \frac{\frac{\sqrt{2}}{2}}{u^2+\frac{1}{2}}\,du\rig ht)-\left(\frac{\sqrt{2}}{4} \int \frac{v}{v^2+\frac{1}{2}}\,dv+ \frac{\sqrt{2}}{4}\int \frac{\frac{\sqrt{2}}{2}}{v^2+\frac{1}{2}}\,dv\rig ht)$
$\displaystyle =\left(\frac{\sqrt{2}}{4}\int \frac{u}{u^2+\frac{1}{2}}\,du - \frac{1}{4}\int \frac{1}{u^2+\frac{1}{2}}\,du\right)-\left(\frac{\sqrt{2}}{4} \int \frac{v}{v^2+\frac{1}{2}}\,dv+ \frac{1}{4}\int \frac{1}{v^2+\frac{1}{2}}\,dv\right)$
$\displaystyle =\frac{\sqrt{2}}{4}\left(\frac{1}{2}\ln(u^2+\frac{ 1}{4})\right)-\frac{1}{4}\left(\frac{1}{\frac{\sqrt{2}}{2}}\tan^ {-1}\frac{u}{\frac{\sqrt{2}}{2}}\right)-\frac{\sqrt{2}}{4}\left(\frac{1}{2}\ln(v^2+\frac{1 }{4})\right)-\frac{1}{4}\left(\frac{1}{\frac{\sqrt{2}}{2}}\tan^ {-1}\frac{v}{\frac{\sqrt{2}}{2}}\right)$
$\displaystyle =\frac{\sqrt{2}}{8}\ln(u^2+\frac{1}{4})-\frac{1}{2\sqrt{2}}\tan^{-1}\frac{2u}{\sqrt{2}}-\frac{\sqrt{2}}{8}\ln(v^2+\frac{1}{4})-\frac{1}{2\sqrt{2}}\tan^{-1}\frac{2v}{\sqrt{2}}$
$\displaystyle =\frac{\sqrt{2}}{8}\ln\left((x+\frac{\sqrt{2}}{2}) ^2+\frac{1}{4}\right)-\frac{1}{2\sqrt{2}}\tan^{-1}\frac{2(x+\frac{\sqrt{2}}{2})}{\sqrt{2}}$
$\displaystyle \ \ -\frac{\sqrt{2}}{8}\ln\left((x-\frac{\sqrt{2}}{2})^2+\frac{1}{4}\right)-\frac{1}{2\sqrt{2}}\tan^{-1}\frac{2(x-\frac{\sqrt{2}}{2})}{\sqrt{2}}$
$\displaystyle =\frac{\sqrt{2}}{8}\ln\left((x+\frac{\sqrt{2}}{2}) ^2+\frac{1}{4}\right)-\frac{1}{2\sqrt{2}}\tan^{-1}{(\sqrt{2}x+1)}$
$\displaystyle \ \ -\frac{\sqrt{2}}{8}\ln\left((x-\frac{\sqrt{2}}{2})^2+\frac{1}{4}\right)-\frac{1}{2\sqrt{2}}\tan^{-1}{(\sqrt{2}x-1)}$
$\displaystyle +\ C$
Now I'm going to collapse and sleep for a while.
Workin' with partial fractions it's just a mess, but possible.
Here's a specific calculation in this case.
$\displaystyle \int {\frac{{x^2 }}
{{x^4 + 1}}\,dx} = \frac{1}
{2}\int {\frac{{\left( {x^2 + 1} \right) + \left( {x^2 - 1} \right)}}
{{x^4 + 1}}\,dx} = \frac{1}
{2}\left[ {\int {\frac{{1 + \dfrac{1}
{{x^2 }}}}
{{x^2 + \dfrac{1}
{{x^2 }}}}\,dx} + \int {\frac{{1 - \dfrac{1}
{{x^2 }}}}
{{x^2 + \dfrac{1}
{{x^2 }}}}\,dx} } \right].$
Now
$\displaystyle \int {\frac{{1 + \dfrac{1}
{{x^2 }}}}
{{x^2 + \dfrac{1}
{{x^2 }}}}\,dx} = \int {\frac{{\left( {x - \dfrac{1}
{x}} \right)'}}
{{\left( {x - \dfrac{1}
{x}} \right)^2 + 2}}\,dx} = \frac{1}
{{\sqrt 2 }}\arctan \frac{{x^2 - 1}}
{{\sqrt 2 x}}.$
$\displaystyle \int {\frac{{1 - \dfrac{1}
{{x^2 }}}}
{{x^2 + \dfrac{1}
{{x^2 }}}}\,dx} = \int {\frac{{\left( {x + \dfrac{1}
{x}} \right)'}}
{{\left( {x + \dfrac{1}
{x}} \right)^2 - 2}}\,dx} = \frac{1}
{{2\sqrt 2 }}\ln \left| {\frac{{x^2 - \sqrt 2 x + 1}}
{{x^2 + \sqrt 2 x + 1}}} \right|.$
Addin' the constant of integration we happily get
$\displaystyle \int {\frac{{x^2 }}
{{x^4 + 1}}\,dx} = \frac{1}
{2}\left[ {\frac{1}
{{\sqrt 2 }}\arctan \frac{{x^2 - 1}}
{{\sqrt 2 x}} + \frac{1}
{{2\sqrt 2 }}\ln \left| {\frac{{x^2 - \sqrt 2 x + 1}}
{{x^2 + \sqrt 2 x + 1}}} \right|} \right] + k.$
I find it fascinating that the two of these are different only by a constant.
One question: Krizalid's solution will not work if the domain of integration includes 0 (or does it?). Why does DivideBy0's solution work for such a region?
-Dan
Well, it's not that fascinating really. The two log terms are easily combined and the result simplified to almost get Krizalid's log term. The two arctan terms are less but not too less easily combined and simplified to almost get Krizalid's arctan term.
The reason for the "almost" is that there are a couple of typos(?) in DivideBy0's answer. Eg 2x + 1 rather than $\displaystyle \sqrt{2}x + 1$ etx. .....