Partial fractions integral. Compute $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$

Question

Evaluate $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$.

I directly approached this with partial fractions and rewritten $x^6+1=(x^2)^3+1=(x^2+1)(x^4+x^2+1)$.

Therefore the integral is:

$$\int_{0}^{1} \frac {x^4+1}{(x^2+1)(x^4+x^2+1)}dx=-2\int_{0}^{1} \frac {x}{1+x^2}dx+2\int_{0}^{1}\frac 1{1+x^2}dx-\int_{0}^1\frac {(x-1)^2}{(x^2+1)^2+(\frac {\sqrt{3}}2)^2}dx$$

But I don't know how to solve the last term.

I answered the question... I realized my silly mistake, sorry for bothering...

Answer 1

I'm sorry but I had a really stupid mistake and now I solved the integral really easily...

So this is what I did:

$$I=\int_0^1 \frac {x^4+1}{(x^2)^3+1}dx=\int_0^1\frac{(x^2+1)^2-2x^2}{(x^2+1)(x^4-x^2+1)}dx$$

then we divide top and bottom by $x^2$ and separate the integrals... we get:

$$I=\int_0^1\frac{1+\frac 1{x^2}}{(x-\frac 1x)^2+1}dx-2\int_0^1\frac{x^2}{(x^3)^2+1}dx$$

and this is:

$$I=arctan(x-\frac 1x)-\frac 23arctan(x^3)$$

which from $0$ to $1$, we get: $I=\frac \pi3.$

Answer 2

Set $x^2=y$ $$\dfrac{y^2+1}{y^3+1}=\dfrac A{y+1}+\dfrac{By+C}{y^2-y+1}$$

$$y^2=y^2(A+B)+y(B+C-A)+A+C$$

$\implies A+C=1\iff C=1-A, B+C=A\implies B=A-C=2A-1,$

$1=A+B=3A-1\iff A=\dfrac23, B=\dfrac13,C=\dfrac13$

$$\implies\dfrac{x^4+1}{x^6+1}=\dfrac1{3(x^2+1)}+\dfrac{x^2+1}{3(x^4-x^2+1)}$$

Now $\dfrac{x^2+1}{x^4-x^2+1}=\dfrac{1+\dfrac1{x^2}}{\left(x-\dfrac1x\right)^2+1}$

set $x-\dfrac1x=u$

Answer 3

Note that $x^6+1=(x^2)^3+1=(x^2+1)(x^4-x^2+1)$ and use the fact that you are integrating over $[0,1]$, $$\int_0^1 \frac{x^4+1\pm x^2}{x^6+1}dx =\int_0^1 \frac{x^4-x^2+1}{x^6+1}dx+ \frac{1}{3}\int_0^1 \frac{3x^2}{x^6+1}dx\\=\int_0^1 \frac{1}{x^2+1}dx+ \frac{1}{3}\int_0^1 \frac{1}{t^2+1}dt=\left(1+\frac{1}{3}\right)\frac{\pi}{4}=\frac{\pi}{3}$$ where, in the second integral, $t=x^3$.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}{x^{4} + 1 \over x^{6} + 1}\,\dd x & = \int_{0}^{1}{x^{4} + 1 - x^{10} - x^{6} \over 1 - x^{12}}\,\dd x = {1 \over 12}\int_{0}^{1}{x^{-7/12} + x^{-11/12} - x^{-1/12} - x^{-5/12} \over 1 - x}\,\dd x \\[5mm] & = {1 \over 12}\pars{\int_{0}^{1}{1 - x^{-5/12} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{-7/12} \over 1 - x}\,\dd x} \\[2mm] & + {1 \over 12}\pars{\int_{0}^{1}{1 - x^{-1/12} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{-11/12} \over 1 - x}\,\dd x} \\[5mm] & = {\pars{H_{-5/12} - H_{-7/12}} + \pars{H_{-1/12} - H_{-11/12}}\over 12}\qquad\pars{~H_{z}:\ Harmonic\ Number~} \\[5mm] & = {\pi\cot\pars{5\pi/12} + \pi\cot\pars{\pi/12}\over 12} \qquad\pars{~Euler\ Reflection\ Formula~} \\[5mm] & = {\pi \over 12}\,{\sin\pars{\pi/2} \over \sin\pars{5\pi/12}\sin\pars{\pi/12}} = {\pi \over 12}\,{1 \over \bracks{\cos\pars{\pi/3} - \cos\pars{\pi/2}}/2} = {\pi \over 6}\,{1 \over 1/2 - 0} \\[5mm] & = \bbx{\pi \over 3} \approx 1.0472 \end{align}