how to solve this integral

Calculus
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How can I solve this ? ∫ (4/x^4+1) dx
Jan 15th, 2015

4/(x4+1)

4/x2 (x2+1/x2)

4/x2 = (2+2/x2)-(2-2/x2)

Integral 2(1+1/x2) /(x2+1/x2)dx- Integral 2(1-1/x2) /(x2+1/x2)dx

In the first integral put x-1/x = t

In the second integral put x+1/x = u

Integral 2 dt /(t2+2)-integral 2 du /(u2-2)

2/√2 tan -1  t/√2- integral 2/2√2 [1/(u-√2)-1/(u+√2)]

2/√2 tan -1  t/√2-1/√2 ln [(u-√2)/(u+√2)]+c

2/√2 tan -1 (x-1/x)/ √2- 1/√2 ln [(x+1/x-√2)/(x+1+√2)]+c


Jan 15th, 2015

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