Question

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

Answer 1

Factorize the denominator then apply partial fraction decomposition.

Explanation:

Let

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

Complete the square in the denominator:

`I=intdx/((x^2+1)^2-2x^2)^2`

Apply the difference of squares:

`I=intdx/((x^2+sqrt2x+1)^2(x^2-sqrt2x+1)^2)`

Apply partial fraction decomposition:

`I=1/(8sqrt2)int{(2x+sqrt2)/(x^2+sqrt2x+1)^2-(2x-sqrt2)/(x^2-sqrt2x+1)^2+(3(x+sqrt2))/(x^2+sqrt2x+1)-(3(x-sqrt2))/(x^2-sqrt2x+1)}dx`

Rearrange:

`I=1/(8sqrt2)int{(2x+sqrt2)/(x^2+sqrt2x+1)^2-(2x-sqrt2)/(x^2-sqrt2x+1)^2+3/2(2x+sqrt2)/(x^2+sqrt2x+1)-3/2(2x-sqrt2)/(x^2-sqrt2x+1)+3/2 sqrt2/(x^2+sqrt2x+1)+3/2 sqrt2/(x^2-sqrt2x+1)}dx`

Complete the square in the denominator of the last two terms:

`I=1/(8sqrt2)int{(2x+sqrt2)/(x^2+sqrt2x+1)^2-(2x-sqrt2)/(x^2-sqrt2x+1)^2+3/2(2x+sqrt2)/(x^2+sqrt2x+1)-3/2(2x-sqrt2)/(x^2-sqrt2x+1)+(3sqrt2)/((sqrt2x+1)^2+1)+(3sqrt2)/((sqrt2x-1)^2+1)}dx`

Integrate term by term:

`I=1/(8sqrt2){-1/(x^2+sqrt2x+1)+1/(x^2-sqrt2x+1)+3/2ln|x^2+sqrt2x+1|-3/2ln|x^2-sqrt2x+1|+3tan^-1(sqrt2x+1)+3tan^-1(sqrt2x-1)}`

Simplify:

`I=1/(8sqrt2){(2sqrt2)/(x^4+1)+3/2ln|(x^2+sqrt2x+1)/(x^2-sqrt2x+1)|+3tan^-1((sqrt2x)/(1-x^2))}`