Vtyjkyuk

I am trying to evaluate the integral

$$int frac11+x^4 mathrm dx.$$

The integrand $frac11+x^4$ is a rational function (quotient of two polynomials), so I could solve the integral if I can find the partial fraction of $frac11+x^4$. But I failed to factorize $1+x^4$.

Any other methods are also wellcome.

Without using fractional decomposition:
beginalign*intfrac1x^4+1 dx & =frac12intfrac21+x^4 dx\ &=frac12intfrac(1-x^2)+(1+x^2)1+x^4 dx\ &=frac12intfrac1-x^21+x^4 dx+frac12intfrac1+x^21+x^4 dx\
&=frac12intfrac1-frac1x^2x^2+frac1x^2, dx+frac12intfrac1+frac1x^2x^2+frac1x^2, dx\ &=frac12left(intfrac1-frac1x^2left(x+frac1xright)^2-2, dx+intfrac1+frac1x^2left(x-frac1xright)^2+2, dxright)\ &=frac12left(intfracdleft(x+frac1xright)left(x+frac1xright)^2-2+intfracdleft(x-frac1xright)left(x-frac1xright)^2+2right)endalign*

So, finally solution is $$intfrac1x^4+1 dx =frac12left(fracarctanleft(fracx^2-1xsqrt2right)sqrt2-frac12sqrt2logleft(fracx^2-sqrt2x+1x^2+sqrt2x+1right)right)+C$$

Hint:

$$x^4+1=(x^2-sqrt2x+1)(x^2+sqrt2x+1) tag1$$
You can integrate using partial fraction decomposition. Since
$$x^4+1=(x^2-sqrt2x+1)(x^2+sqrt2x+1),$$
then
$$frac1x^4+1=fracAx+Bx^2-sqrt2x+1+fracCx+Dx^2+sqrt2x+1=frac(Ax+B)(x^2+sqrt2x+1)+(Cx+D)(x^2-sqrt2x+1)(x^2-sqrt2x+1)(x^2+sqrt2x+1) = \
= fracx^3(A+C)+x^2(Asqrt2+B+D-Csqrt2)+x(Bsqrt2-Dsqrt2)+B+D(x^2-sqrt2x+1)(x^2+sqrt2x+1)$$
$$begincases
A+C=0;\
B+D+sqrt2(A-C)=0; \
B-D=0; \
B+D=1.
endcases$$

$$
A=-C=-frac12sqrt2; \
B=D=frac12$$

$$
frac1x^4+1=frac12sqrt2left(dfrac-x+sqrt2x^2-sqrt2x+1+ dfracx+sqrt2x^2+sqrt2x+1 right).
$$

Added

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

1. Completion to the full square
   $$x^4+1=x^4 +2x^2+1 -2x^2=(x^2+1)^2-left(sqrt2xright)^2 = \ =big(x^2-sqrt2x +1 big)big( x^2+sqrt2x +1 big).$$




2. Let $omega_i, iin1,, 2,,3,,4 $ are roots of the equation $x^4+1=0$ over $mathbbC:$ $omega_1=fracsqrt22(1-i), omega_2=fracsqrt22(1+i) omega_3=fracsqrt22(-1+i), omega_4=fracsqrt22(-1-i). $ Then use the decomposition into prime factors and multiply pairwise complex conjugate:
   $x^4+1= left( x-fracsqrt22(1-i) right)left( x-fracsqrt22(1+i) right)left( x-fracsqrt22(-1+i) right)left( x-fracsqrt22(-1-i) right) = \ =big(x^2-sqrt2x +1 big)big( x^2+sqrt2x +1 big).$

$$I =int frac1x^4+1 dx$$

If we add and subtract $2x^2$ to $x^4 + 1$, we get:
$$int frac1x^4 + 2x^2 + 1 - 2x^2$$

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

$$int frac1(x^2 + 1)^2 - 2x^2$$

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

Hence, $(x^2 + 1)^2 - 2x^2 = (x^2 + 1 - sqrt2x)(x^2 + 1 + sqrt2x)$

$$int frac1(x^2-sqrt2x+1)(x^2+sqrt2x+1)$$

Now using partial fraction decomposition:

$$frac1(x^2-sqrt2x+1)(x^2+sqrt2x+1) = fracAx + Bx^2-sqrt2x+1 + fracCx + Dx^2+sqrt2x+1$$

After you have found A, B, C and D, its just a basic $fraclinearquadratic$ type integral.

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

$$frac1x^4+1=frac ax-omega_0+frac bx-omega_1+frac cx-omega_2+frac dx-omega_3$$

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

$$a=lim_xtoomega_0fracx-omega_0x^4+1=frac14omega_0^3,$$
and similarly for the other terms.

Then, a single term is integrated with a complex logarithm
$$intfracdxx-omega=ln(x-omega)=ln|x-omega|+iangle(x-omega).$$

Here we have $omega_0=dfrac1+isqrt2$, hence

$$lnsqrt(x-frac1sqrt2)^2+(frac1sqrt2)^2-iarctanfracfrac1sqrt2x-frac1sqrt2\
=frac12ln(x^2-sqrt2x+1)-ifracpi2+iarctan(sqrt2x-1).$$

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