Evaluating $\int\frac{x^4+1}{x^6+1}dx$

Question

I have problem with this integral: $$\int\dfrac{x^4+1}{x^6+1}dx$$ I guess it is easy, but I was trying for quite a long time and the only thing I got is headache.

Thanks for help

Hint: $x^6 + 1 = (x^2+1)(x^4 - x^2 + 1), x^4 + 1= (x^4 - x^2 + 1) + (x^2 + 1) - 1$ — GohP.iHan, Apr 21 '15 at 17:17
@copper.hat Then I have $\int \frac{x^4}{x^6+1}+\int \frac{1}{x^6+1}$. The second integral should be equal to $\arctan x^3$ and the first one? — GorTeX, Apr 21 '15 at 17:20
Also factorize $x^4 - x^2 + 1 = (x^2 - \sqrt{3} x + 1)(x^2 + \sqrt{3} x + 1)$. — Robert Israel, Apr 21 '15 at 17:25
$\frac{x^4+1}{x^6+1}=\frac{1}{3}(\frac{x^2+1}{x^4-x^2+1}+\frac{2}{x^2+1})$, then please read this http://math.stackexchange.com/questions/928040/how-to-solve-the-integral-int-frac-x2-1x4-x2-1-dx — MonkeyKing, Apr 21 '15 at 17:26

score 3 · Accepted Answer · answered Apr 21 '15 at 17:29

3

Try $$ \frac{x^4+1}{x^6+1}=\frac{x^4-x^2+1}{x^6+1}+\frac{x^2}{x^6+1} =\frac1{x^2+1}+\frac{x^2}{x^6+1}. $$ The first term is a table integral. The second term reduces to it with the substitution $t=x^3, dt=3x^2\,dx$.

answered Apr 21 '15 at 17:29

Jyrki Lahtonen

133,153

Thanks :-) this helped a lot – GorTeX Apr 21 '15 at 17:34
This yields $\arctan x+\dfrac13 \arctan(x^3)$ for the primitive. Admittedly it is not entirely clear to me why that equals $\dfrac23 \arctan x-\dfrac13\arctan\dfrac x{x^2-1}, $ which is what Mathematica gives as the answer. Both of those vanish at $x=0$, so there is no difference in the "$+C$" part :-). – Jyrki Lahtonen Apr 21 '15 at 17:42

score 0 · Answer 2 · edited Apr 21 '15 at 18:01

0

hint

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

And proceed to the next step in decomposing the second fraction of the RHS

edited Apr 21 '15 at 18:01

abiessu

8,115

answered Apr 21 '15 at 17:26

marwalix

16,773

score 0 · Answer 3 · answered Apr 21 '15 at 19:11

$\bf{My\; Solution:: }$ Let $$\displaystyle I = \int\frac{x^4+1}{x^6+1}dx = \int \frac{(x^2+1)^2-2x^2}{(x^2+1)\cdot (x^4-x^2+1)}dx$$

$$\displaystyle I = \int\frac{x^2+1}{x^4-x^2+1}dx-2\int\frac{x^2}{(x^3)^2+1}dx = \int\frac{\left(1+\frac{1}{x^2}\right)}{\left(x-\frac{1}{x}\right)^2+1}dx-\frac{2}{3}\int\frac{3x^2}{(x^3)^2+1}dx$$

For First Integral put $$\displaystyle \left(x-\frac{1}{x}\right)=t\;\;,$$ Then $$\left(1+\frac{1}{x^2}\right)dx =dt$$

And for Second Integral Put $$x^3=u\;\;, 3x^2dx=du$$

So We Get $$\displaystyle I = \int\frac{x^4+1}{x^6+1}dx = \tan^{-1}\left(x-\frac{1}{x}\right)-\frac{2}{3}\tan^{-1}(x^3)+\mathcal{C}$$

Evaluating $\int\frac{x^4+1}{x^6+1}dx$

3 Answers3