Question

How will you integrate ? `int(dx)/(1+x^4)^2`

Answer 1

`intdx/(x^4+1)^2`

= `intx^(-8)dx/(x^(-4)+1)^2`

= `1/4*intx^(-3)*4x^(-5)*dx/[x^(-4)+1]^2`

= `1/4*x^(-3)*(x^(-4)+1)^(-1)-1/4*int(-3)*x^(-4)*dx/[x^(-4)+1]`

= `1/4*x^(-3)/(x^(-4)+1)+3/4*intx^(-4)*dx/[x^(-4)+1]`

= `1/4*x/(x^4+1)+3/4*intdx/(x^4+1)`

= `1/4*x/(x^4+1)+3/8*int2*dx/(x^4+1)`

= `1/4*x/(x^4+1)+3/8*int2x^(-2)*dx/(x^2+x^(-2))`

= `1/4*x/(x^4+1)+3/8*int(1+x^(-2))*dx/(x^2+x^(-2))-3/8*int(1-x^(-2))*dx/(x^2+x^(-2))`

= `1/4*x/(x^4+1)+3/8*int(1+x^(-2))*dx/[(x-x^(-1))^2+2)-3/8*int(1-x^(-2))*dx/((x+x^(-1))^2-2)`

= `1/4*x/(x^4+1)+3/16*Sqrt(2)*Arctan[(x-x^(-1))/(2)^(1/2)]-3/16*Sqrt(2)*Arcsinh[(x+x^(-1))/(2)^(1/2)]+C`

= `1/4*x/(x^4+1)+3/16*Sqrt(2)*Arctan[(x^2-1)/(x*Sqrt(2))]-3/16*Sqrt(2)*Arcsinh[(x^2+1)/(x*Sqrt(2))]+C`

Explanation:

1) I divided denominator and numerator of integrand with x^8

2) I decomposed numerator for resembling derivative of denominator.

3) I used partial fraction

4) I expanded fractions with x^4

5) I started to decompose second integral by multiply and divide with 2

6) I divided denominator and numerator of second integrand with x^2

7) I decomposed second one for resembling denominators of them at forms of `u^2+a^2` for u=`x-x^(-1)` and `u^2-a^2` for u=`x+x^(-1)`

8) I integrated decomposed them.

9) I rewrote results.