How to solve the integral $\int \frac {(x^2 +1)}{x^4- x^2 +1} dx$

Question

I have started this problem but I'm not completely sure I'm going down the right path with it.

So far I have completed the square in the denominator.

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

Then, let $u=x^2-\frac{1}{2}$ so $x=\sqrt(u+\frac{1}{2})$

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

If you let $u=x^2-\frac{1}{2}$ then what is $\frac{du}{dx}$? — Hassan Muhammad, Sep 11 '14 at 21:26

score 5 · Answer 1 · answered Sep 12 '14 at 05:21

5

$$\frac{x^2+1}{x^4-x^2+1}=\frac{1+\dfrac1{x^2}}{x^2-1+\dfrac1{x^2}}$$

Now $\displaystyle\int\left(1+\dfrac1{x^2}\right)dx=x-\dfrac1x$

and $\displaystyle x^2-1+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2-1$

Hope you can take it from here

answered Sep 12 '14 at 05:21

lab bhattacharjee

274,582

@user175401, How about this? – lab bhattacharjee Sep 12 '14 at 05:21
thats a perfect method – Ekaveera Gouribhatla Sep 12 '14 at 05:33

score 4 · Answer 2 · answered Sep 11 '14 at 21:13

4

Split it :

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

Then remember that $\arctan(x)'=\frac{1}{1+x^2}$

answered Sep 11 '14 at 21:13

Hippalectryon

7,730

a sixth power in that denom ??? – imranfat Sep 11 '14 at 21:15
@imranfat Why is that a problem ? – Hippalectryon Sep 11 '14 at 21:16
I think it's just surprising. This is an extremely slick method. – preferred_anon Sep 11 '14 at 21:17
Your post wasn't finished when I read it, for some reason, the fraction after arctan did not appear on the screen, so I thought things were typed wrong... – imranfat Sep 11 '14 at 21:20
@imranfat yeah, there was a typo I fixed on the fly on that line – Hippalectryon Sep 11 '14 at 21:21
Hello, did you use partial fractions to split the fraction? – JohnK Sep 11 '14 at 22:20
@JohnK I didn't need to, I just knew (because I had encountered it before) that we could split it like that. Had I not know that beforehand, I would have indeed used partial fractions. – Hippalectryon Sep 11 '14 at 22:21

score 1 · Answer 3 · answered Sep 12 '14 at 05:41

As Hassan Muhammad commented, you have a mistake. Since you set $x=\sqrt{u+\frac{1}{2}}$, you then have $dx=\frac{1}{2 \sqrt{u+\frac{1}{2}}}$ and then $$I=\int \frac {(x^2 +1)}{x^4- x^2 +1} dx=\int \frac{2 u+3}{\sqrt{u+\frac{1}{2}} \left(4 u^2+3\right)}du$$ which is not very nice.

If you use Hippalectryon's idea, you end with $$I=\int \frac {(x^2 +1)}{x^4- x^2 +1} dx=\int\left(\frac{3 x^{2}}{x^{6} + 1} + \frac{1}{x^{2} + 1}\right)dx=\tan ^{-1}\left(x^3\right)+\tan ^{-1}\left(x\right)$$ and if you apply the formula $$\tan ^{-1}\left(a\right)+\tan ^{-1}\left(b\right)=\tan ^{-1}\left(\frac{a+b}{1-a b}\right)$$ you have, after simplifications, $$I=\tan ^{-1}\left(\frac{x}{1-x^2}\right)$$ which is result to which lab bhattacharjee's answer is taking you to.

It's simpler, at least for me, to use $$u=x2−1/2\du=2xdx\ \frac{du}{2x}=dx$$ but the result is the same. — nickalh, Sep 12 '14 at 07:05
@nickalh. There are many ways to do it, I agree ! As long as we arrive to the result, they are fine. Cheers :-) — Claude Leibovici, Sep 12 '14 at 07:11

How to solve the integral $\int \frac {(x^2 +1)}{x^4- x^2 +1} dx$

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