Evaluating $\int \frac{t}{1+t^2+t^4+t^8}dt$ I am stuck to evaluate  the following integral in terms of finite terms (with out using power series technique). 
$$\int  \frac{t}{1+t^2+t^4+t^8}dt$$
Could anyone help me?
 A: First, change variables to $u = t^2$, because we can and 8th order polynomials are awful.
$$
\int \frac{t\,dt}{1+t^2+t^4+t^8} = \frac{1}{2}\int\frac{du}{1+u+u^2+u^4}
$$
Next, since $1+u+u^2+u^4$ has no real zeroes and no $u^3$ term, there exist $a,b,c\in\mathbb R^+$ such that $1+u+u^2+u^4 = [(u+a)^2 + b^2][(u-a)^2+c^2]$. These numbers have a very complicated closed form, but their numeric values are $a \approx 0.547$,  $b \approx 0.586$, $c \approx 1.121$.
Next comes partial fraction time. There are four real numbers $A,B,C,D$ such that
\begin{multline}
\frac{1}{[(u-a)^2 + b^2][(u+a)^2+c^2]} = \frac{Ab}{[(u+a)^2 + b^2]} + \frac{Bc}{[(u-a)^2 + c^2]} \\ + \frac{2C(u+a)}{[(u+a)^2 + b^2]}  + \frac{2D(u-a)}{[(u-a)^2 + c^2]}
\end{multline}
This turns out to have the solution
$$
A = \frac{4a^2-b^2 + c^2}{b\Delta}\;\;\;;\;\;\;B = \frac{4a^2 + b^2 - c^2}{c\Delta}\;\;\;;\;\;\; C = -D = \frac{2a}{\Delta}
$$
where $\Delta = (b^2-c^2)^2 + 8a^2(b^2+c^2) + 16a^4$. These numbers also have a very complicated closed form, and their numeric values are $A \approx 0.0417$, $B \approx 0.591$, $C = -D\approx0.179$.
Lastly, each term in the partial fraction can be integrated using the formulae
$$
\int \frac{b\, du}{(u+a)^2 +b^2}  =\tan^{-1}\left(\frac{u+a}{b}\right)\;\;\;\;;\;\;\;\;\int\frac{2(u+a)du}{(u+a)^2+b^2} = \ln\left[(u+a)^2+b^2\right]
$$
Doing this and using $u = t^2$ again gives us the final antiderivative:
$$
\int\frac{tdt}{1+t^2+t^4+t^8} = \frac{1}{2}\left(A\tan^{-1}\left[\frac{t^2-a}{b}\right] + B\tan^{-1}\left[\frac{t^2+a}{c}\right] + C\ln\left[\frac{(t^2+a)^2+b^2}{(t^2-a)^2+c^2}\right]\right)
$$
You're not really going to do any better than this. The roots of the polynomial in the denominator have no simple form.
A: It's a series, but a finite series
It can be done, but it isn't simple. One must take a theoretical approach here. If we consider the polynomial $H(x)=1+x^2+x^4+x^8$, and it's set of roots $R=\{x\in\Bbb C:H(x)=0\}$, then we can factor $H(x)$ as a product over it's roots as follows:
$$H(x)=F\prod_{r\in R}(x-r)$$
Where $F$ is some constant. With this we can perform fraction decomposition on $\frac1{H(x)}$. We start with
$$\frac1{H(x)}=\frac1F\prod_{r\in R}\frac1{x-r}$$
Then to decompose the fraction we say that 
$$\prod_{r\in R}\frac1{x-r}=\sum_{r\in R}\frac{b(r)}{x-r}$$
Where $b(r)$ is some constant (different for each $r$) that is independent of $x$. We continue:
$$\bigg(\prod_{a\in R}(x-a)\bigg)\bigg(\prod_{r\in R}\frac1{x-r}\bigg)=\sum_{r\in R}\frac{b(r)}{x-r}\prod_{a\in R}(x-a)$$
The LHS reduces to $1$, and the RHS simplifies:
$$1=\sum_{r\in R}b(r)\prod_{r\neq a\in R}(x-a)$$
Thus for any $k\in R$, 
$$1=\sum_{r\in R}b(r)\prod_{r\neq a\in R}(k-a)$$
Which gives 
$$1=b(k)\prod_{k\neq a\in R}(k-a)$$
$$b(k)=\prod_{k\neq a\in R}\frac1{k-a}$$
Thus we can write our integral as 
$$I=\frac1F\int\sum_{r\in R}\frac1{x-r}\prod_{r\neq a\in R}\frac1{r-a}\ \mathrm{d}x$$
$$I=\frac1F\sum_{r\in R}\prod_{r\neq a\in R}\frac1{r-a}\int\frac{\mathrm{d}x}{x-r}$$
$$I=\frac1F\sum_{r\in R}\log|x-r|\prod_{r\neq a\in R}\frac1{r-a}\ +C$$
Unfortunately I do not know how to find $F$ or closed forms of the roots. If I do I'll include it.
