How will you integrate ?

#int(dx)/(1+x^4)^2#

Question

How will you integrate ?

#int(dx)/(1+x^4)^2#

Emma Aldridge · Accepted Answer

Factorize the denominator then apply partial fraction decomposition.

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))}#

Charles Anctil · Answer

undefined To integrate $ \int \frac{dx}{(1+x^4)^2} $, we can use the substitution method. Let $ u = x^2 $ and $ du = 2x \, dx $.

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

Now, we can use partial fraction decomposition to integrate $ \frac{1}{(1+u^2)^2} $. Let's represent $ \frac{1}{(1+u^2)^2} $ as $ \frac{A}{1+u^2} + \frac{Bu+C}{(1+u^2)^2} $.

By solving for $ A $, $ B $, and $ C $, we integrate each term separately and then substitute $ u = x^2 $ back in to obtain the final result.