I need help solving the indefinite integral of 1/(1+x^4). Do I use Tan^-1 or log?

Answer 1

Integral dx / ( 1 + x^4 )
= ( 1 /2 ) Integral [ ( 1 + x^2 ) / ( 1 + x^4 ) + ( 1 - x^2 ) / ( 1 + x^4 ) ]dx
= (1/2) Integral [ (1/x^2 + 1)/(1/x^2 + x^2) + (1/x^2 - 1)/(1/x^2 + x^2)]dx
= (1/2) Integral [ (1+1/x^2)/((x - 1/x)^2 + 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
Hence, (1 + 1/x^2) dx = du and ( 1 - 1/x^2 ) dx = dv

So, Integral = (1/2) Integral [ du/(u^2 + 2) - dv/(v^2 - 2) ]

=1/(2sqrt2)tan inverse (u/sqrt2) - 1/(4sqrt2) log l(v - sqrt2)/(v + sqrt2)l
Substitute for u and v to get the final answer.

Answer 2

Looks like you need both. This integral looks like something well beyond a first year calculus class.

Here is a link.

http://integrals.wolfram.com/index.jsp