integrate: (x²-1) / [(x⁴+3x²+1) * tan^-1{(x²+1)/x}] dx

${\int }_{}$(x²-1) / [(x⁴+3x²+1) * tan^-1{(x²+1)/x}] dx
Divide numerator and denominator ay x² ,
We get  ${\int }_{}$ (1-1/x²) / [{(x+1/x)²+1}* tan^-1(x + (1/x))] ,
Put  tan^-1(x+(1/x)) = t and diffrentiate on both side we get,

⇒ (1 - 1/x²) dx / {1+(x+1/x)²} = dt
Now substituting this value then term 1+(x+1/x)² will get cancelled and finally we will left with , 

 ⇒ ∫ 1/t. dt = logt + C = log tan^-1(x+1/x) + C