$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$ $$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$$
I tried to solve it.
$$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x^3+x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x(x^2-1)+x^2+1}dx$$
But this does not seem to be solving.Please help.
 A: Using Hermite reduction and the Rothstein-Trager Algorithm for integration of rational functions I get
$$\frac{2}{\sqrt{11}}\left(\arctan\left(\frac{3}{\sqrt{11}}+\frac{2}{\sqrt{11}}x\right)-\arctan\left(\frac{3}{\sqrt{11}}-\frac{8}{\sqrt{11}}x-\frac{6}{\sqrt{11}}x^2-\frac{2}{\sqrt{11}}x^3\right)\right)
$$
which seems correct after differentiating. Its strange that Wolfram Alpha doesn't get this result.
Here some details:
Let $A = x^2+1$ and $D = x^4+3x^3+3x^2-3x+1$, so that we want to find $\int{\frac{A}{D}}{dx}$
Since $D$ is square free (contains no multiple factors) and $\deg(A) < \deg(D)$, we can apply the Rothstein-Trager method described in the paper below to compute this integral:
First we compute the resultant 
\begin{align}
\text{res}_x(D, A-t\frac{d}{dx}D) &= \text{res}_x(x^4+3x^3+3x^2-3x+1,-4tx^3+(1-9t)x^2-6tx+3t+1)\\ &= 37(1+11t)^2 =: R(t)
\end{align}
It has two distinct roots $t_1 = \frac{i}{\sqrt{11}}$ and $t_2 = -\frac{i}{\sqrt{11}} = -t_1$ 
We now can express the integral as a sum of complex logarithms:
\begin{align}
\int{\frac{A}{D}}{dx} &= \sum_{t|R(t) = 0}t\log\left(\gcd\left(D, A-t\frac{d}{dx}D\right)\right)\\ &= \sum_{t|R(t) = 0}t\log\left(-1+\left(\frac{3}{2}+\frac{11}{2}t\right)x+x^2\right) \\ &= \frac{i}{\sqrt{11}}\log\left(-1+\left(\frac{3}{2}+\frac{\sqrt{11}}{2}i\right)x+x^2\right)-\frac{i}{\sqrt{11}}\log\left(-1+\left(\frac{3}{2}-\frac{\sqrt{11}}{2}i\right)x+x^2\right)
\end{align}
Note that the GCD inside the logs was taken over $\mathbb{Q}(t_1)[x]$, rather than just over $\mathbb{Q}[x]$.
Having expressed the integral as a sum of logarithms, the only thing that we have to do, is to convert them to inverse tangents. For this, there also exist algorithms. 
See for this "Symbolic Integration I: Transcendental Functions" by Manuel Bronstein.
A: Don't take this answer too seriously, but I could not resist. I leave details to you to fill in.
First, you factor the denominator into second degree factors, as
$$
\begin{aligned}
x^4+3x^3+3x^2-3x+1&=\bigl(x^2+\frac{1}{2} \sqrt{7+2 \sqrt{37}} x+\frac{3 x}{2}+\frac{1}{4} \sqrt{46+10 \sqrt{37}}+\frac{\sqrt{37}}{4}+\frac{5}{4}\bigr)\\
&\qquad\times\bigl(x^2-\frac{1}{2} \sqrt{7+2 \sqrt{37}} x+\frac{3 x}{2}-\frac{1}{4} \sqrt{46+10 \sqrt{37}}+\frac{\sqrt{37}}{4}+\frac{5}{4}\bigr)
\end{aligned}
$$
Then, do a partial fraction decomposition. The ansatz is
$$
\begin{aligned}
\frac{x^2+1}{x^4+3x^3+3x^2-3x+1}&=\frac{ax+b}{x^2+\frac{1}{2} \sqrt{7+2 \sqrt{37}} x+\frac{3 x}{2}+\frac{1}{4} \sqrt{46+10 \sqrt{37}}+\frac{\sqrt{37}}{4}+\frac{5}{4}}\\
&\qquad+\frac{cx+d}{x^2-\frac{1}{2} \sqrt{7+2 \sqrt{37}} x+\frac{3 x}{2}-\frac{1}{4} \sqrt{46+10 \sqrt{37}}+\frac{\sqrt{37}}{4}+\frac{5}{4}}.
\end{aligned}
$$
A direct calculation shows that $a=c=0$ and
$$
b=\frac{1}{2}+\sqrt{\frac{\sqrt{37}}{22}-\frac{7}{44}}
\quad
\text{and}
\quad
d=\frac{1}{2}-\sqrt{\frac{\sqrt{37}}{22}-\frac{7}{44}}.
$$
What you then have to do is the usual completing the square in the denominator, and you will get some horrible arctans. I leave that to you.
A: Please note that you have: $\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$, which on dividing numerator and denominator by $x^2$ becomes:
$\int\frac{1+1/x^2}{(x^2+1/x^2)+3+3(x-1/x)}dx$=$\int\frac{1+1/x^2}{(x-1/x)^2+5+3(x-1/x)}dx$
Now put x-1/x =t so that (1+1/$x^2$)dx=dt and thus you get:
$\int\frac{1}{t^2+5+3t}dt$;which can be rearranged to get:$\int\frac{1}{t^2+5+3t}dt$=$\int\frac{1}{(t+3/2)^2+11/4}dt$=(2/$\sqrt11$)arctan[(t+3/2)2/$\sqrt11$]+c=(2/$\sqrt11$)arctan[(x-1/x+3/2)2/$\sqrt11$]+c;where c is integration constant.
PS:x>$0$ or x<$0$,only for these x , above solution is valid.At,x=$0$,arctan[(x-1/x+3/2)2/$\sqrt11$]becomes discontinuous so Fundamental theorem doesn't hold.
A: First rewrite the integral as
$$\mathcal{I}\stackrel{def}{=}\int \frac{x^2+1}{x^4 + 3x^3 + 3x^2 - 3x + 1} dx
= \int \frac{x+x^{-1}}{(x^2 + x^{-2}) + 3(x - x^{-1}) + 3}\frac{dx}{x}
$$
Using the identity: $\displaystyle\;\frac{dx}{x} = \frac{d(x-x^{-1})}{x+x^{-1}}\;$,
we can change variable to $u = x-x^{-1}$ and get
$$
\mathcal{I} 
= \int\frac{du}{u^2 + 3u + 5} = \int\frac{du}{(u+\frac32)^2 + \frac{11}{4}}
= \sqrt{\frac{4}{11}}
\tan^{-1}\left(\sqrt{\frac{4}{11}}\left(u+\frac32\right)\right) + C\\
= \frac{2}{\sqrt{11}}
\tan^{-1}\left(\frac{2}{\sqrt{11}}\left(x - \frac1x +\frac32\right)\right) + C
$$
for some integration constant $C$.
Update
As pointed out by @Andrei, the expression above is discontinuous at $x = 0$. If one want to use it to compute definite integral, one need to use a different $C$
for the region $x > 0$ and $x < 0$. Let $C_{+}$ and $C_{-}$ be the integration constant for this two regions. Since the integrand is well behaved at $x = 0$,
as a function of $x$, the indefinite integral $\mathcal{I}$ is continuous at $x = 0$. This impose a constraint
$$C_{+} - C_{-} = \frac{2\pi}{\sqrt{11}}$$
and leads to
$$\mathcal{I} = \frac{2}{\sqrt{11}}
\left[ 
\tan^{-1}\left(\frac{2}{\sqrt{11}}\left(x - \frac1x +\frac32\right)\right) +
\Delta(x)
\right] + C'
$$
where
$\displaystyle\;C' = \frac{C_{+} + C_{-}}{2}\;$
 and $\displaystyle\;
\Delta(x) = \begin{cases}+\frac{\pi}{2}, & x > 0\\-\frac{\pi}{2},& x < 0\end{cases}$.
To further simplify this, we use following representation of $\Delta(x)$:
$$\Delta(x) = \tan^{-1}x + \tan^{-1}\frac1x,\quad\text{ for } x \ne 0$$
This leads to
$$\begin{align}
\mathcal{I} 
&= \frac{2}{\sqrt{11}}
\left[ 
\tan^{-1}\left(\frac{2}{\sqrt{11}}\left(x - \frac1x +\frac32\right)\right) +
\tan^{-1}\left(\frac{2}{\sqrt{11}x}\right)
+ \tan^{-1}\left(\frac{\sqrt{11}x}{2}\right)
\right] + C'\\
&= \frac{2}{\sqrt{11}}
\left[
\tan^{-1}\left(\frac{\sqrt{11}x^2(2x+3)}{7x^2-6x+4}\right)
+ \tan^{-1}\left(\frac{\sqrt{11}x}{2}\right)
\right] + C'
\end{align}
$$
An expression of $\mathcal{I}$ which is continuous for all $x$ and one can use to compute definite integral.
A: As already pointed out, the roots of the polynomial are complex. 
The analytic solving is a boring task. Only the main intermediate results are reported below :

A: Writing the polynomial $x^4+3x^3+3x^2-3x+1$ as a product of two polynomials of degree two, i.e,. $$x^4+3x^3+3x^2-3x+1= (x^2+Ax+B)(x^2+Cx+D)$$, then we will get 4 equations $A+C=3,B+D+AC=3,AD+BC=-3,BD=1$, using wolframalpha.com to solve this eqs. we get 
Thus, $$\int{\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx}=\int{\frac{1+x^2}{(x^2+Ax+B)(x^2+Cx+D)}dx}$$
Using partial fraction, we get $$\frac{1+x^2}{(x^2+Ax+B)(x^2+Cx+D)}=\frac{Px+Q}{x^2+Ax+B}+\frac{Tx+S}{x^2+Cx+D}$$
Solving for $P,Q,T$ and $S$ ...etc
Therefore, 
$$\int{\frac{1+x^2}{(x^2+Ax+B)(x^2+Cx+D)}dx}=\int{\frac{Px+Q}{x^2+Ax+B}dx}+\int{\frac{Tx+S}{x^2+Cx+D}dx}=\text{constant}_1 \ln|x^2+Ax+B|+\text{constant}_2 \ln|x^2+Cx+D|+ CONSTANT$$
