how to solve $\int \frac{x^2+1}{x^4-x^2+1}dx$

Question

I saw this problem $$\int \frac{x^2+1}{x^4-x^2+1}dx$$ I checked wolfram and the answer is $$\arctan\left(\frac{x}{1-x^2}\right)+C$$ but I want a method of the solution so it can help me to solve similar problems

Answer 1

$$I = \int \frac{x^{2} + 1}{x^{4} - x^{2} + 1} dx = \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}} - 1} dx$$

Let $u = x - \frac{1}{x}$ then $du = (1 + \frac{1}{x^{2}}) dx$ and

$$\begin{aligned} I = \int \frac{1}{u^{2} + 1} du &= \arctan(u) + C \\ &= \arctan \left( \frac{x^{2} - 1}{x} \right) + C \end{aligned}$$

Using the fact that $\arctan(x) + \arctan \left( \frac{1}{x} \right) = \frac{\pi}{2}$, we can check that it matches Wolfram's answer. Using partial fractions is possible, but it is very dirty and not recommended since the denominator has only non-real zeros.

Answer 2

Beside the elegant solution given by @Vue, just write $$\frac {x^2+1}{x^4-x^2+1}=\frac {x^2+1}{(x^2-a)(x^2-b)}$$ Now, using partial fraction decomposition, the integrand is just $$\frac 1{a-b}\left(\frac{a+1}{x^2-a}-\frac{b+1}{x^2-b}\right)$$ and integration leads to $$\frac 1{a-b}\left(\frac{(b+1) \tanh ^{-1}\left(\frac{x}{\sqrt{b}}\right)}{\sqrt{b}}-\frac{(a+1) \tanh ^{-1}\left(\frac{x}{\sqrt{a}}\right)}{\sqrt{a}}\right)$$

Using now the fact that $$a=\frac{1+i \sqrt{3}}{2}\qquad \text{and} \qquad b=\frac{1-i \sqrt{3}}{2}$$ replace, simplify the complex numbers to obtain $$\int\frac {x^2+1}{x^4-x^2+1}\,dx=\tan ^{-1}\left(\frac{x}{2} \left(1-i \sqrt{3} \right)\right)+\tan ^{-1}\left(\frac{x}{2} \left(1+i \sqrt{3} \right)\right)$$ and now $$\tan^{-1}(A)+\tan^{-1}(B)=\cdots\cdots$$

The advantage of this is that you can apply it to any polynomial denominator.

Answer 3

Hint: Substitute $x=\sqrt{\dfrac{1-y}{1+y}}$ to get a simpler integral,

$$\begin{align*} &\int \frac{x^2+1}{x^4-x^2+1} \, dx \\ &= - \int \frac{\frac{1-y}{1+y}+1}{\frac{(1-y)^2}{(1+y)^2}-\frac{1-y}{1+y}+1} \, \sqrt{\frac{1+y}{1-y}} \, \frac{dy}{(1+y)^2} \\ &= -2 \int \frac{dy}{\left(3y^2+1\right)\sqrt{1-y^2}} \end{align*}$$

Answer 4

I decided to visit my old questions to see If I could solve them now

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

let $x=\tan(t) $ $dx = \sec^2(t) dt $ $$I = \int \frac{\sec^2(t)}{\sec^2(t) -3\left( 1-\frac{1}{\sec^2(t)}\right) } dt= \int \frac{\sec^2(t)}{\sec^2(t) -3 \sin^2(t) } dt=\int \frac{1}{1 - \frac{3}{4}\sin^2(2t) } dt$$ $$ = \int \frac{1}{ \cos^2(2t)+ \frac{1}{4}\sin^2(2t) } dt= \int \frac{2* \frac{1}{2}\sec^2(2t)}{ 1+ \frac{1}{4}\tan^2(2t) } dt = \arctan \left( \frac{\tan(2t)}{2} \right) +C $$

since $\tan(2t) =\frac{2 \tan(t)}{ 1-\tan^2(t)} $ and since $tan(t)=x$ then $\tan(2t) =\frac{2 x}{ 1-x^2} $

so that $$I =\arctan \left( \frac{\tan(2t)}{2} \right) +C =\arctan\left( \frac{ \tan(t)}{ 1-\tan^2(t)}\right) +C=\arctan\left(\frac{ x}{ 1-x^2} \right)+C $$