How will you integrate ? int(dx)/(1+x^4)^2∫dx(1+x4)2
1 Answer
Factorize the denominator then apply partial fraction decomposition.
Explanation:
Let
I=intdx/(1+x^4)^2I=∫dx(1+x4)2
Complete the square in the denominator:
I=intdx/((x^2+1)^2-2x^2)^2I=∫dx((x2+1)2−2x2)2
Apply the difference of squares:
I=intdx/((x^2+sqrt2x+1)^2(x^2-sqrt2x+1)^2)I=∫dx(x2+√2x+1)2(x2−√2x+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)}dxI=18√2∫⎧⎪ ⎪⎨⎪ ⎪⎩2x+√2(x2+√2x+1)2−2x−√2(x2−√2x+1)2+3(x+√2)x2+√2x+1−3(x−√2)x2−√2x+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))}
