How to evaluate $\int_0^1\frac{1+x^4}{1+x^6}\,dx$

Question

$$\int_0^1\frac{1+x^4}{1+x^6}\,dx$$

Can anyone help me solve the question? I am struggling with this.

RE a previous edit: How are the tags unnecessary? The tags are applicable. Also, rolling back keeps the integral in an off-site image and also keeps the title vague---both of which should be avoided. — Chris Brooks, Mar 25 '14 at 02:07
@Joseph: Yep. Seems like people just click approve without even reading it. I would have rejected it too, but it was quickly approved. — user2345215, Mar 25 '14 at 02:12

score 13 · Accepted Answer · 2014-11-19T05:12:57.553

13

$$ \begin{align} \int_{0}^{1} \frac{x^{4}+1}{x^{6}+1} \ dx &= \int_{0}^{1} \frac{x^{4}+1+x^{2}-x^{2}}{x^{6}+1} \ dx \\ &= \int_{0}^{1} \frac{x^{4}-x^{2}+1}{x^{6}+1} \ dx + \int_{0}^{1}\frac{x^{2}}{x^{6}+1} \ dx \\ &= \int_{0}^{1} \frac{x^{4}-x^{2}+1}{(x^{2}+1)(x^{4}-x^{2}+1)} \ dx + \int_{0}^{1}\frac{x^{2}}{(x^{3})^{2}+1} \ dx \\ &= \int_{0}^{1} \frac{1}{x^{2}+1} \ dx + \frac{1}{3} \int_{0}^{1} \frac{1}{u^{2}+1} \ du \\ &= \frac{4}{3} \int_{0}^{1} \frac{1}{x^{2}+1} \ dx \\&= \frac{4}{3} \left(\frac{\pi}{4} \right) \\ &= \frac{\pi}{3} \end{align}$$

edited Nov 19 '14 at 05:12

answered Mar 25 '14 at 02:34

That's fantastic! THANK YOU! – python3 Mar 25 '14 at 02:37
another question: Can you tell me how to enter some symbols of mathematics in the website if I want to ask some questions? I am new here! – python3 Mar 25 '14 at 02:39
@tiandiao123 here is a link with an overview of mathjax commands. – Chris Brooks Mar 25 '14 at 03:52
Nice algebra show, RV! – Mathemagician1234 Nov 19 '14 at 05:45
converted to CW at author's request. – robjohn Nov 19 '14 at 08:32
@tiandiao123 Please don't. Just ignore RV's wish. This is a nice answer & no one will accuse him of copying someone else answer. :-) – Venus Nov 20 '14 at 09:16

score 5 · Answer 2 · answered Nov 19 '14 at 05:35

RV's great answer notwithstanding, here's another way using the residue theorem. Note that

$$\int_0^1 dx \frac{1+x^4}{1+x^6} = \frac14 \int_{-\infty}^{\infty} dx \frac{1+x^4}{1+x^6}$$

The integral on the RHS is, by the residue theorem, $i 2 \pi$ times the sum of the residues of the poles of the integrand in the upper half plane, or

$$\begin{align}i \frac{\pi}{2} \left (\frac{1+e^{i 4 \pi/6}}{6 e^{i 5 \pi/6}} + \frac{2}{6 i}+\frac{1+e^{i 20 \pi/6}}{6 e^{i 25 \pi/6}}\right ) &= i\frac{\pi}{6} \left (-i + e^{-i \pi/6} + e^{-i 5 \pi/6} \right )\\ &= \frac{\pi}{6} \left ( 1+2 \sin{\frac{\pi}{6}}\right ) \\ &= \frac{\pi}{3}\end{align}$$

Whoa,that's even better from a mathematical point of view-never thought of it! — Mathemagician1234, Nov 19 '14 at 05:47
That's similar to the way I was about to answer, using this answer with $m=6$ and then $n=0$ and $n=4$. (+1) — robjohn, Nov 19 '14 at 07:53

How to evaluate $\int_0^1\frac{1+x^4}{1+x^6}\,dx$

2 Answers2