integrate ((x^2+1)/ (x^4+4))dx

Question

integrate ((x^2+1)/ (x^4+4))dx

Answer 1

I don't know where you get this problem but this makes me use all of my calculus knowledge.

 

First, we need to split to manageable partial fractions:

(x²+1)/(x⁴+1) = (-1/8x+1/4)/(x²+2x+2) + (1/8x+1/4)/(x²-2x+2)

 

Then, integrate the first half and the second half separately (for easiness)

∫(-1/8x+1/4)/(x²+2x+2) dx = (-1/8)∫(x-2)/(x²+2x+2) dx = (-1/16)∫(2x+2)/(x²+2x+2) - 6/(x²+2x+2) dx = (-1/16)[ln(x²+2x+2)- 6tan^-1(x+1)]+C

Similarly,

∫(1/8x+1/4)/(x²-2x+2) dx = (1/8)∫(x+2)/(x²-2x+2) dx = (1/16)∫(2x-2)/(x²-2x+2) + 6/(x²-2x+2) dx = (1/16)[ln(x²-2x+2)+6tan^-1(x-1)] +C

Combining both answers and simplify, they become (-1/16)[ln(x²+2x+2)- 6tan^-1(x+1)]+(1/16)[ln(x²-2x+2)+6tan^-1(x-1)] +C = (1/16)(ln((x²-2x+2)/(x²+2x+2))+6tan^-1(x+1)+6tan^-1(x-1)) +C