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(1+x²)/(1+x⁴)

  divide numerator and denominator by x²

(1+1/x²)/(1/x²+x²)

(1+1/x²⁾/[(x-1/x)² +2]    as x²+1/x² = (x-1/x)² +2

take (x-1/x)=t

on differentiating

dt=(1+1/x² )dx

now

we hav integral of dt/(t² +2)

1/(2)^1/2  tan^-1t/(2)^1/2 +c   substitute value of t in above and u will get answer

Her solution is correct,but you could also use the completion of squares method for this problem.the problem can be solved by making the denominator a pefect square and you automatically get the dx term(for the derivative) in the numerator.from then on it is plain simple integration of inverse of tan.happy solving!

1+x²/1+x⁴ dx

=1+1/x²/x²+1/x² dx

in denominator add 2 & subtract 2

1+1/x²/(x-1/x)²+2 dx

let x-1/x =t

= 1+1/x² dx =dt

 =dt/t²+2^1/2

= 1/2^1/2tan^-1t/2^1/2

=1/2^1/2tan^-1x-1/x/2^1/2