integration

Note that:

x4 + 1 = (x2 + 1)2 - 2x2

= (x2 + x√2 + 1)(x2 - x√2 + 1)

Hence 1/(x4 + 1) can be expressed in partial fraction form:

(Ax + B)/(x2 + x√2 + 1) + (Cx + D)/(x2 - x√2 + 1)

where A, B, C and D are constants.

Then:

(Ax + B)(x2 - x√2 + 1) + (Cx + D)(x2 + x√2 + 1) = 1

(A + C)x3 + (-A√2 + B + C√2 + D)x2 + (A - B√2 + C + D√2)x + (B + D) = 1

Comparing the coefficients on both sides, we have:

A + C = 0, -A√2 + B + C√2 + D = 0, A - B√2 + C + D√2 = 0 and B + D = 1

Solving:

A = 1/(2√2), B = 1/2, C = -1/(2√2), D = 1/2

Hence

1/(x4 + 1) = [1/(2√2)] [(x + √2)/(x2 + x√2 + 1) - (x - √2)/(x2 - x√2 + 1)] ∫dx/(x4 + 1) = [1/(2√2)] [∫(x + √2)dx/(x2 + x√2 + 1) - ∫(x - √2)dx/(x2 - x√2 + 1)] For ∫(x + √2)dx/(x2 + x√2 + 1):∫(x + √2)dx/(x2 + x√2 + 1) = (1/2) ∫(2x + 2√2)dx/(x2 + x√2 + 1)= (1/2) ∫d(x2 + x√2 + 1)/(x2 + x√2 + 1) + (1/√2) ∫dx/(x2 + x√2 + 1)= (1/2) ln (x2 + x√2 + 1) + (1/√2) ∫dx/[(x + 1/√2)2 + 1/2]= (1/2) ln (x2 + x√2 + 1) + (1/√2) ∫dx/[(x + 1/√2)2 + (1/√2)2]= (1/2) ln (x2 + x√2 + 1) + tan-1 [√2(x + 1/√2)] + C= (1/2) ln (x2 + x√2 + 1) + tan-1 (√2x + 1) + CFor ∫(x - √2)dx/(x2 - x√2 + 1):∫(x - √2)dx/(x2 - x√2 + 1) = (1/2) ∫(2x - 2√2)dx/(x2 - x√2 + 1)= (1/2) ∫d(x2 - x√2 + 1)/(x2 + x√2 + 1) - (1/√2) ∫dx/(x2 - x√2 + 1)= (1/2) ln (x2 - x√2 + 1) - (1/√2) ∫dx/[(x - 1/√2)2 + 1/2]= (1/2) ln (x2 - x√2 + 1) - (1/√2) ∫dx/[(x - 1/√2)2 + (1/√2)2]= (1/2) ln (x2 - x√2 + 1) - tan-1 [√2(x - 1/√2)] + C= (1/2) ln (x2 - x√2 + 1) - tan-1 (√2x - 1) + CFinally:∫dx/(x4 + 1) = [1/(2√2)] [(1/2) ln (x2 + x√2 + 1) + tan-1 (√2x + 1) - (1/2) ln (x2 - x√2 + 1) + tan-1 (√2x - 1)] + C= [1/(2√2)] {(1/2) ln [(x2 + x√2 + 1)/(x2 - x√2 + 1)] + tan-1 (√2x + 1) + tan-1 (√2x - 1)} + C= [1/(4√2)] ln [(x2 + x√2 + 1)/(x2 - x√2 + 1)] + [1/(2√2)] [tan-1 (√2x + 1) + tan-1 (√2x - 1)] + CFor ∫dx/(x5 + 1), pls. refer to: http://hk.knowledge.yahoo.com/question/question?qid=7007052605338

The steps are so complicated that I don't think anyone will answer these 2 questions, suggest you refer to WolframAlpha for details.