# Integral with binomial to a power $\int\frac{1}{(x^4+1)^2}dx$

I have to solve the following integral:

$$\int\frac{1}{(x^4+1)^2}dx$$

I tried expanding it and then by partial fractions but I ended with a ton of terms and messed up. I also tried getting the roots of the binomial for the partial fractions but I got complex roots and got stuck. Is there a trick for this kind of integral or some kind of helpful substitution? Thanks.

EDIT:

I did the following:

Let $$x^2=\tan\theta$$, then $$x = \sqrt{\tan\theta}$$ and $$dx=\frac{\sec^2\theta}{2x}d\theta$$

Then:

$$I=\int\frac{1}{(x^4+1)^2}dx = \int\frac{1}{(\tan^2\theta+1)^2} \frac{\sec^2\theta}{2x}d\theta=\int\frac{1}{\sec^4\theta} \frac{\sec^2\theta}{2x}d\theta$$

$$I=\frac{1}{2}\int{\frac{1}{\sec^2\theta \sqrt{\tan\theta}}}d\theta$$.

After this I don't know how to proceed.

• Have you tried a trigonometric substitution? This type of substitution is common in integrals that have this form. – Rebeca Lie Yatsuzuka Silva Feb 6 '20 at 20:57
• I thought that there must be a useful trigonometric substitution but I can't figure out which one. – Rolando González Feb 6 '20 at 21:10
• Maybe with tangent, secant and co-secant? – Rebeca Lie Yatsuzuka Silva Feb 6 '20 at 21:18

Use $$\left(\frac x{x^4+1}\right)' = -\frac3{x^4+1} + \frac 4{(x^4+1)^2}$$ to rewrite the integral as

$$I = \int \frac 1{(x^4+1)^2}dx=\frac x{4(x^4+1)}+\frac34\int\frac1{x^4+1} dx$$

where the integral on the RHS is

$$\int\frac2{x^4+1} dx = \int\frac{1+x^2}{x^4+1} dx + \int\frac{1-x^2}{x^4+1} dx$$ $$= \int\frac{\frac1{x^2}+1}{x^2+\frac1{x^2}} dx + \int\frac{\frac1{x^2}-1}{x^2+\frac1{x^2}} dx = \int\frac{d(x-\frac1{x})}{(x-\frac1{x})^2+2} - \int\frac{d(x+\frac1{x})}{(x+\frac1{x})^2-2}$$ $$=\frac1{\sqrt2} \tan^{-1}\frac{x^2-1}{\sqrt2x} + \frac1{\sqrt2} \coth^{-1}\frac{x^2+1}{\sqrt2x}$$

Thus,

$$I = \frac x{4(x^4+1)}+\frac3{8\sqrt2} \tan^{-1}\frac{x^2-1}{\sqrt2x} + \frac3{8\sqrt2} \coth^{-1}\frac{x^2+1}{\sqrt2x} + C$$

• Quick check so I think your approach was brilliant but your arithmetic must have gone wrong early on. – user5713492 Feb 7 '20 at 7:39
• @user5713492 - Appreciate the note and fix it. – Quanto Feb 7 '20 at 12:53
• OK, but now what do you get for $$\int_{-\infty}^{\infty}\frac{dx}{(x^4+1)^2}$$? – user5713492 Feb 7 '20 at 16:10
• @user5713492 - you get $2\int_0^\infty \frac 1{(x^4+1)^2}dx = \frac{3\pi}{4\sqrt2}$ – Quanto Feb 7 '20 at 17:53
• @Isham - Note $\coth^{-1} x = \frac12 \ln \frac {x+1}{x-1}$, which is a concise way to express the result – Quanto Feb 9 '20 at 15:26

I am not aware of a trick. I would just write $$x^4+1$$ as $$\left(x^2+\sqrt2x+1\right)\left(x^2-\sqrt2x+1\right)$$ and then I would write$$\frac1{(x^4+1)^2}$$as$$\frac{Ax+B}{x^2+\sqrt2x+1}+\frac{Cx+D}{\left(x^2+\sqrt2x+1\right)^2}+\frac{Ex+F}{x^2-\sqrt2x+1}+\frac{Gx+H}{\left(x^2-\sqrt2x+1\right)^2}.$$

One way to go is to expand into linear factors. Let $$\omega_k=\exp(\pi i(2k+1)/4)$$, so $$\omega_k^4=-1$$ and $$\frac1{(x^4+1)^4}=\sum_{k=0}^3\left(\frac{A_k}{(x-\omega_k)^2}+\frac{B_k}{x-\omega_k}\right)$$ Then $$A_k=\lim_{x\rightarrow\omega_k}\frac{(x-\omega_k)^2}{(x^4+1)^2}=\left(\lim_{x\rightarrow\omega_k}\frac{x-\omega_k}{x^4+1}\right)^2=\left(\frac1{4\omega_k^3}\right)^2=\left(\frac{-\omega_k}{4}\right)^2=\frac{\omega_k^2}{16}$$ and \begin{align}B_k&=\lim_{x\rightarrow\omega_k}\frac d{dx}\frac{(x-\omega_k)^2}{(x^4+1)^2}\\ &=\lim_{x\rightarrow\omega_k}2\frac{(x-\omega_k)}{(x^4+1)}\frac{\left(x^4+1-4x^3(x-\omega_k)\right)}{(x^4+1)^2}\\ &=2\left(\frac{-\omega_k}4\right)\lim_{x\rightarrow\omega_k}\frac{-12x^2(x-\omega_k)}{8x^3(x^4+1)}\\ &=2\left(\frac{-\omega_k}4\right)\left(\frac{-3}{2\omega_k}\right)\left(\frac{-\omega_k}4\right)=\frac{-3\omega_k}{16}\end{align} So now \begin{align}\int\frac{dx}{(x^4+1)^2}&=\frac1{16}\sum_{k=0}^3\int\left(\frac{\omega_k^2}{(x-\omega_k)^2}-\frac{3\omega_k}{x-\omega_k}\right)dx\\ &=\frac1{16}\sum_{k=0}^3\left(\frac{-\omega_k^2}{x-\omega_k}-3\omega_k\ln(x-\omega_k)\right)+C\end{align} Now, $$\omega_{3-k}=\omega_k^*$$ and $$\frac{-\omega_k^2}{x-\omega_k}+\frac{-\left(\omega_k^*\right)^2}{x-\omega_k^*}=\frac{-\left(\omega_k^2+\left(\omega_k^*\right)^2\right)x+\omega_k+\omega_k^*}{x^2-\left(\omega_k+\omega_k^*\right)x+1}=\frac{2\cos\frac{\pi(2k+1)}{4}}{x^2-2x\cos\frac{\pi(2k+1)}{4}+1}$$ Also \begin{align}-\omega_k\ln(x-\omega_k)-\omega_k^*\ln(x-\omega_k^*)&=-\frac12(\omega_k+\omega_k^*)\left(\ln(x-\omega_k)+\ln(x-\omega_k^*)\right)\\ &\quad-\frac12(\omega_k-\omega_k^*)\left(\ln(x-\omega_k)-\ln(x-\omega_k^*)\right)\\ &=-\cos\frac{\pi(2k+1)}4\ln\left(x^2-2x\cos\frac{\pi(2k+1)}4+1\right)\\ &\quad-i\sin\frac{\pi(2k+1)}4\left(-2i\tan^{-1}\left(\frac{\sin\frac{\pi(2k+1)}4}{x-\cos\frac{\pi(2k+1)}4}\right)\right)\end{align} So that \begin{align}\int\frac{dx}{(x^4+1)^2}&=\frac1{16}\left\{\frac{\sqrt2}{x^2-\sqrt2\,x+1}-\frac{\sqrt2}{x^2+\sqrt2\,x+1}\right.\\ &\quad-\frac3{\sqrt2}\ln\left(x^2-\sqrt2\,x+1\right)+\frac3{\sqrt2}\ln\left(x^2+\sqrt2\,x+1\right)\\ &\quad\left.-3\sqrt2\tan^{-1}\left(\frac1{\sqrt2\,x-1}\right)-3\sqrt2\tan^{-1}\left(\frac1{\sqrt2\,x+1}\right)\right\}+C\\ &=\frac x{4{(x^4+1)}}+\frac3{16\sqrt2}\ln\left(\frac{x^2+\sqrt2\,x+1}{x^2-\sqrt2\,x+1}\right)-\frac{3\sqrt2}{16}\tan^{-1}\left(\frac{\sqrt2\,x}{x^2-1}\right)+C\end{align} Quick check

EDIT: There is a problem with the above expression in that it is discontinuous when $$x=\pm1$$. To fix this, note that \begin{align}\tan^{-1}y&=2\tan^{-1}\left(\tan\frac12\tan^{-1}y\right)=2\tan^{-1}\left(-\frac1y+\sqrt{\frac1{y^2}-1}\right)\\ &=2\tan^{-1}\frac{\sqrt2\,x}{\sqrt{x^4+1}-x^2+1}\end{align} For the angle we taking taking inverse tangent of above, so \begin{align}\int\frac{dx}{(x^4+1)^2}&=\frac x{4{(x^4+1)}}+\frac3{16\sqrt2}\ln\left(\frac{x^2+\sqrt2\,x+1}{x^2-\sqrt2\,x+1}\right)\\ &\quad-\frac{3}{4\sqrt2}\tan^{-1}\left(\frac{\sqrt2\,x}{\sqrt{x^4+1}+x^2-1}\right)+C\end{align} Check again

EDIT: I was trying so hard to avoid the discontinuity at $$x=0$$ that I made it even worse. I should have gone with \begin{align}\tan^{-1}\left(\frac1{\sqrt2\,x-1}\right)+\tan^{-1}\left(\frac1{\sqrt2\,x+1}\right)&=\tan^{-1}\left(\frac{\sqrt2\,x}{x^2-1}\right)\\ &=2\tan^{-1}\left(\frac{-1+\sqrt{1+\left(\frac{\sqrt2\,x}{x^2-1}\right)^2}}{\left(\frac{\sqrt2\,x}{x^2-1}\right)}\right)\\ &=2\tan^{-1}\left(\frac{-x^2+1-\sqrt{x^4+1}}{\sqrt2\,x}\right)\\ &=2\tan^{-1}\left(\left(-\frac x{\sqrt2}\right)\left(1+\frac{x^2}{\sqrt{x^4+1}}\right)\right)\end{align} Where I would finally have gotten rid of all discontinuities at $$x\in\{-1,0,1\}$$ or stayed with $$\tan^{-1}\left(\frac1{\sqrt2\,x-1}\right)+\tan^{-1}\left(\frac1{\sqrt2\,x+1}\right)=-\tan^{-1}(\sqrt2\,x-1)-\tan^{-1}(\sqrt2\,x+1)$$ and avoided the combination of arctangents entirely. By avoiding the discontinuities I can get an expression that evaluates $$\int_{-\infty}^{\infty}\frac{dx}{(x^4+1)^2}=\frac{3\pi\sqrt2}8$$ correctly.

Hints:

$$\frac1{(x^4+1)^2}=\frac{x^4+1-x^4}{(x^4+1)^2}=\frac1{x^4+1}-\frac{x^4}{(x^4+1)^2}$$ and by parts

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

This way we can get rid of the square at the denominator, and we are left with

$$\frac1{x^4+1}.$$

Now using the factorization of the quartic binomial,

$$\frac{\sqrt8}{x^4+1}=\frac{x+\sqrt2}{x^2+\sqrt2x+1}-\frac{x-\sqrt2}{x^2-\sqrt2x+1}.$$

Here, by completing the square, we can handle the terms $$\sqrt2x$$ in the denominators, and solve with terms $$\log(x^2\pm\sqrt2x+1)$$ and $$\arctan(\sqrt2x\pm1)$$.