Q.252. Integration

Question 252.
Find ∫ (1/((x^4)+1)) dx.

Answer 252.
∫ (1/((x^4)+1)) dx
= (1/2) ∫ [ (x^2+1)/(x^4+1) - (x^2-1)/(x^4+1) ] dx
= (1/2) ∫(1 + 1/x^2)/(x^2 + 1/x^2) dx - (1/2) ∫ (1 - 1/x^2)/(x^2 + 1/x^2) dx
= (1/2) ∫(1+1/x^2)/[(x-1/x)^2+2] dx - (1/2) ∫(1-1/x^2)/[(x+1/x)^2-2] dx

Let x-1/x = u for the first integral and x+1/x = v for the second integral
=> (1+1/x^2) dx = du and (1-1/x^2) dx = dv

=> Integral
= (1/2) ∫ du/(u^2+2) - (1/2) ∫dv/(v^2-2)
= 1/(2√2) tanֿ¹u/√2 - (1/4√2) ln l (v - √2)/(v + √2) l + c
= 1/(2√2) tanֿ¹ (x-1/x)/√2 - (1/4√2) ln l (x+1/x - √2)/(x+1/x + √2) l + c
= 1/(2√2) tanֿ¹ (x^2-1)/√2x - (1/4√2) ln l (x^2-√2 x+1)/(x^2+√2 x+1) + c.

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