This question already has an answer here:

I am trying to evaluate the integral

$$\int \frac{1}{1+x^4} \mathrm dx.$$

The integrand $\frac{1}{1+x^4}$ is a rational function (quotient of two polynomials), so I could solve the integral if I can find the partial fraction of $\frac{1}{1+x^4}$. But I failed to factorize $1+x^4$.

Any other methods are also wellcome.


marked as duplicate by Nosrati, Lord Shark the Unknown, Saad, Cesareo, MR_BD Oct 3 '18 at 8:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I suppose it's been long enough (6 to 7 weeks). At this Math Forum archived post you'll find 3 .pdf files giving extremely detailed evaluations of the antiderivative of $\frac{1}{x^n + 1}$ for $n=4,\;5,$ and $6.$ $\endgroup$ – Dave L. Renfro May 7 '13 at 19:43

Without using fractional decomposition: \begin{align*}\int\frac{1}{x^4+1}\ dx & =\frac{1}{2}\int\frac{2}{1+x^{4}}\ dx\\\ &=\frac{1}{2}\int\frac{(1-x^{2})+(1+x^{2})}{1+x^{4}}\ dx\\\ &=\frac{1}{2}\int\frac{1-x^2}{1+x^{4}}\ dx+\frac{1}{2}\int\frac{1+x^{2}}{1+x^{4}}\ dx\\ &=\frac{1}{2}\int\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}\, dx+\frac{1}{2}\int\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}\, dx\\ &=\frac{1}{2}\left(\int\frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-2}\, dx+\int\frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+2}\, dx\right)\\ &=\frac{1}{2}\left(\int\frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^2-2}+\int\frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+2}\right)\end{align*}

So, finally solution is $$\int\frac{1}{x^4+1}\ dx =\frac{1}{2}\left(\frac{\arctan\left(\frac{x^2-1}{x\sqrt{2}}\right)}{\sqrt{2}}-\frac{1}{2\sqrt{2}}\log\left(\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right)\right)+C$$

  • $\begingroup$ This is really difficult . You are genius . $\endgroup$ – SmartCoder Mar 1 '18 at 13:49

$$x^4+1=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1) \tag{1}$$ You can integrate using partial fraction decomposition. Since $$x^4+1=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1),$$ then $$\frac{1}{x^4+1}=\frac{Ax+B}{x^2-\sqrt{2}x+1}+\frac{Cx+D}{x^2+\sqrt{2}x+1}=\frac{(Ax+B)(x^2+\sqrt{2}x+1)+(Cx+D)(x^2-\sqrt{2}x+1)}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)} = \\ = \frac{x^3(A+C)+x^2(A\sqrt{2}+B+D-C\sqrt{2})+x(B\sqrt{2}-D\sqrt{2})+B+D}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}$$ $$\begin{cases} A+C=0;\\ B+D+\sqrt{2}(A-C)=0; \\ B-D=0; \\ B+D=1. \end{cases}$$

$$ A=-C=-\frac{1}{2\sqrt{2}}; \\ B=D=\frac{1}{2}$$

$$ \frac{1}{x^4+1}=\frac{1}{2\sqrt{2}}\left(\dfrac{-x+\sqrt{2}}{x^2-\sqrt{2}x+1}+ \dfrac{x+\sqrt{2}}{x^2+\sqrt{2}x+1} \right). $$

Decomposition (1) can be done using one of the following ways:

  1. Completion to the full square $$x^4+1=x^4 +2x^2+1 -2x^2=(x^2+1)^2-\left(\sqrt{2}x\right)^2 = \\ =\big(x^2-\sqrt{2}x +1 \big)\big( x^2+\sqrt{2}x +1 \big).$$

  2. Let $\omega_i, \ i\in\{1,\, 2,\,3,\,4\} $ are roots of the equation $x^4+1=0$ over $\mathbb{C}:$ $\omega_1=\frac{\sqrt{2}}{2}(1-i), \ \omega_2=\frac{\sqrt{2}}{2}(1+i) \ \omega_3=\frac{\sqrt{2}}{2}(-1+i), \ \omega_4=\frac{\sqrt{2}}{2}(-1-i). $ Then use the decomposition into prime factors and multiply pairwise complex conjugate: $x^4+1= \left( x-\frac{\sqrt{2}}{2}(1-i) \right)\left( x-\frac{\sqrt{2}}{2}(1+i) \right)\left( x-\frac{\sqrt{2}}{2}(-1+i) \right)\left( x-\frac{\sqrt{2}}{2}(-1-i) \right) = \\ =\big(x^2-\sqrt{2}x +1 \big)\big( x^2+\sqrt{2}x +1 \big).$

  • $\begingroup$ Its too difficult to calculate yet $\endgroup$ – kalpeshmpopat Mar 18 '13 at 8:51
  • 3
    $\begingroup$ He said "FULL ANSWER"! $\endgroup$ – Amihai Zivan Mar 18 '13 at 9:06
  • $\begingroup$ Why is that too difficult? Well, you could do $$\int\frac1{x^4+4}\,dx$$ first using the factorization coming from $$x^4+4=(x^4+4x^2+4)-4x^2=\cdots$$ and then do a linear substitution to get back to your integral. $\endgroup$ – Jyrki Lahtonen Mar 18 '13 at 9:06
  • 13
    $\begingroup$ We won't do your homework - we try to teach you to do it yourself. $\endgroup$ – Jyrki Lahtonen Mar 18 '13 at 9:07
  • $\begingroup$ @M.Strochyk How did you compute $x^4+1$ to be equal to $(x^2-\sqrt{2}x +1)(x^2+\sqrt{2}x+1)$? Is there a general method for doing that? $\endgroup$ – Mr Reality Nov 5 '17 at 5:02

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

If we add and subtract $2x^2$ to $x^4 + 1$, we get: $$\int \frac{1}{{x^4 + 2x^2 + 1 - 2x^2}}$$

We know $x^4 + 2x^2 + 1 = (x^2 + 1)^2$

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

We know $a^2 - b^2 = (a - b)(a + b)$

Hence, $(x^2 + 1)^2 - 2x^2 = (x^2 + 1 - \sqrt{2}x)(x^2 + 1 + \sqrt{2}x)$

$$\int \frac{1}{{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}}$$

Now using partial fraction decomposition:

$$\frac{1}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)} = \frac{Ax + B}{x^2-\sqrt{2}x+1} + \frac{Cx + D}{x^2+\sqrt{2}x+1}$$

After you have found A, B, C and D, its just a basic $\frac{linear}{quadratic}$ type integral.

  • $\begingroup$ Is there a general method for factorizing stuff like $x^4+1$? $\endgroup$ – Mr Reality Nov 5 '17 at 5:11

The indefinite integral is a rational fraction and is typically solved using partial fractions decomposition.

You first factor the denominator $x^4+1$, which has four complex roots, the fourth roots of minus one, let $\omega_0$, $\omega_1$, $\omega_2$, and $\omega_3$. Then you decompose

$$\frac1{x^4+1}=\frac a{x-\omega_0}+\frac b{x-\omega_1}+\frac c{x-\omega_2}+\frac d{x-\omega_3}$$

The unknown coefficients are found by multiplying by one of the denominators and taking the limit to the root:

$$a=\lim_{x\to\omega_0}\frac{x-\omega_0}{x^4+1}=\frac1{4\omega_0^3},$$ and similarly for the other terms.

Then, a single term is integrated with a complex logarithm $$\int\frac{dx}{x-\omega}=\ln(x-\omega)=\ln|x-\omega|+i\angle(x-\omega).$$

Here we have $\omega_0=\dfrac{1+i}{\sqrt2}$, hence

$$\ln\sqrt{(x-\frac1{\sqrt2})^2+(\frac1{\sqrt2})^2}-i\arctan\frac{\frac1{\sqrt2}}{x-\frac1{\sqrt2}}\\ =\frac12\ln(x^2-\sqrt2x+1)-i\frac\pi2+i\arctan(\sqrt2x-1).$$

Repeat for the four terms (there is a lot of symmetry) and form the linear combination.

  • $\begingroup$ I could not understand the answer after Step 1 . $\endgroup$ – user589548 Oct 1 '18 at 15:41

Not the answer you're looking for? Browse other questions tagged or ask your own question.