$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$

Question

$$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$$

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.

Answer 1

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.

Answer 2

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 :

Answer 3

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.

Answer 4

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.

Answer 5

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,

$$\int{\frac{1+x^2}{(x^2+Ax+B)(x^2+Cx+D)}dx}=\int{\frac{Px+Q}{x^2+Ax+B}dx}+\int{\frac{Tx+S}{x^2+Cx+D}dx}=\text{constant}_1 \ln|x^2+Ax+B|+\text{constant}_2 \ln|x^2+Cx+D|+ CONSTANT$$

Answer 6

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.