# Is it hard to evaluate $\int \frac{x^{3}+1}{x^{6}+1} d x$?

Noting that \begin{aligned} \frac{x^{3}+1}{x^{6}+1} &=\frac{x\left(x^{2}+1\right)-(x-1)}{x^{6}+1} =\frac{x}{x^{4}-x^{2}+1}-\frac{x-1}{x^{6}+1} \end{aligned} Therefore

$$\int \frac{x^{3}+1}{x^{6}+1} d x=\underbrace{\int \frac{x d x}{x^{4}-x^{2}+1}}_{H}-\underbrace{\int \frac{x-1}{x^{6}+1} d x}_{M}$$

Let’s start with the easier one $$H$$.

\begin{aligned} H &=\frac{1}{2} \int \frac{d y}{y^{2}-y+1}=\frac{1}{2} \int \frac{d y}{\left(y-\frac{1}{2}\right)^{2}+\frac{3}{4}} \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 y-1}{\sqrt{3}}\right)+c_{1}=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x^{2}-1}{\sqrt{3}}\right)+c_1 \end{aligned} For the harder one $$M$$, we first split the integrand into 2 and then 3.

Let $$\frac{x-1}{x^{6}+1} \equiv \frac{A x+B}{x^{2}+1}+\frac{f(x)}{x^{4}-x^{2}+1}$$ for some constant $$A, B$$ and cubic polynomial $$f(x).$$

Then comparing their numerator yields

$$x-1 \equiv(A x+B)\left(x^{4}-x^{2}+1\right)+\left(x^{2}+1\right) f(x)$$

Putting $$x =i$$ yields

\begin{aligned}i-1 & = (A i+B)(1+1+1) \\&=3 A i+3 B \\ A &=\frac{1}{3} \text { and } B=-\frac{1}{3} \\\left(x^{2}+1\right) f(x) &=x-1-\frac{1}{3}(x-1)\left(x^{4}-x^{2}+1\right) \\f(x) &=-\frac{(x-1)\left(x^{2}-2\right)}{3}\end{aligned} \tag*{}

Putting them back to the integral gives

\displaystyle \begin{aligned}\int \frac{x-1}{x^{6}+1} d x&=\frac{1}{3} \int \frac{x-1}{x^{2}+1} d x-\frac{1}{3} \int \frac{(x-1)\left(x^{2}-2\right)}{x^{4}-x^{2}+1} d x\\&=\frac{1}{3}\left[\underbrace{\int\frac{x-1}{x^{2}+1}-d x}_{J}-\underbrace{\int \frac{x\left(x^{2}-2\right)}{x^{4}-x^{2}+1}}_{K} d x+\underbrace{\int \frac{x^{2}-2}{x^{4}-x^{2}+1}}_{L} d x\right]\end{aligned}\tag*{}

We first start split the integral $$J$$ into two and get $$\displaystyle J=\frac{1}{2} \ln \left(x^{2}+1\right)-\tan ^{-1} x+c_{2}\tag*{}$$

For the integral $$K$$ , letting $$y=x^2$$ yields

\displaystyle \begin{aligned}K &=\frac{1}{2} \int \frac{y-2}{y^{2}-y+1} d y \\&=\frac{1}{2} \int \frac{\frac{1}{2}(2 y-1)-\frac{3}{2}}{y^{2}-y+1} d y \\&=\frac{1}{4} \int \frac{d\left(y^{2}-y+1\right)}{y^{2}-y+1}-\frac{3}{4} \int \frac{d y}{\left(y-\frac{1}{2}\right)^{2}+\frac{3}{4}} \\&=\frac{1}{4} \ln \left(x^{4}-x^{2}+1\right)-\frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{2 x^{2}-1}{\sqrt{3}}\right)+c_{3}\end{aligned}\tag*{}

For the integral $$L$$, let’s play a little trick on the integrand and get

\displaystyle \begin{aligned}L &=\int \frac{x^{2}-2}{x^{4}-x^{2}+1} d x \\&=\int \frac{1-\frac{2}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-1} d x \\&=\int \frac{-\frac{1}{2}\left(1+\frac{1}{x^{2}}\right)+\frac{3}{2}\left(1-\frac{1}{x^{2}}\right)}{x^{2}+\frac{1}{x^{2}}-1} d x\\&=-\frac{1}{2} \int \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+1}+\frac{3}{2} \int \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^{2}-3}\\&=-\frac{1}{2} \tan ^{-1}\left(x-\frac{1}{x}\right)+\frac{\sqrt{3}}{4} \ln \left|\frac{x+\frac{1}{x}-\sqrt{3}}{x+\frac{1}{x}+\sqrt{3}}\right|+c_{4}\end{aligned}\tag*{}

Putting them together concludes that

$$\displaystyle I=\frac{1}{12}\left[-2 \ln \left(x^{2}+1\right)+4 \tan ^{-1} x+\ln \left|x^{4}-x^{2}+1\right|-\frac{5\sqrt3}{3} \tan ^{-1}\left(\frac{2 x^{2}-1}{\sqrt{3}}\right)+2 \tan ^{-1}\left(x-\frac{1}{x}\right)-\sqrt{3} \ln \left|\frac{x^{2}-\sqrt{3} x+1}{x^{2}+\sqrt{3} x+1}\right|\right]+C \tag*{}$$

• So is your question just whether there's a shorter route to that antiderivative? (The shortest proof of its correctness is to differentiate it, but you probably want an approach that finds it, not one that merely verifies it.)
– J.G.
Aug 5 at 16:15
• You are right! I just want to know whether there are simpler methods.
– Lai
Aug 5 at 16:28
• You can actually factor $x^4-x^2+1$ into factors of the form $x^2\pm \sqrt{3}x+1$. This lets you do a nicer partial fraction decomposition. Aug 5 at 16:34
• @xpaul, you are right. Thank you very much for catching what I missed. Fixed now.
– Lai
Aug 5 at 23:45
• The only way faster would be to use a tool like Mathematica...
– PC1
Aug 6 at 0:30

$$\frac{x^3+1}{x^6+1}=\frac{1+i}{-6 x+6 i}-\frac{1-i}{6 x+6 i}+\frac{1+i}{6-3 \left(\sqrt{3}-i\right) x}+$$ $$\frac{1-i}{6+3 \left(\sqrt{3}-i\right) x}+\frac{1+i}{6+3 \left(\sqrt{3}+i\right) x}-\frac{1-i}{-6+3 \left(\sqrt{3}+i\right) x}$$ leads to the sum of six logarithms to be recombined as three logarithms and three arctangents.
• I like this one. And it works in the case $x$ is a complex variable, unlike the ones with $\log|\dots|$ in them. Aug 6 at 10:25
Here is an approach via symmetry. Let $$I=\int\frac{1+x^3}{1+x^6}dx\>\> \>\>\>\>\>J=\int\frac{x+x^4}{1+x^6}dx$$ Then, show that \begin{align} I-J =&\ \frac1{\sqrt3}\tanh^{-1}\frac{\sqrt3x}{1+x^2}+\frac12\ln\frac{\sqrt[3]{1+x^6}}{1+x^2}\\ I+J =&\ \frac23\tan^{-1}x+\frac13\tan^{-1}\frac x{1-x^2}+\frac1{\sqrt3}\tan^{-1}\frac{\sqrt3}{1-2x^2}\\ \end{align}