∫[-1,0][(3x^4+3x^2+1)/(1+x^2)]dx
=∫[-1,0] 3x^2dx +∫[-1,0] 1/(1+x^2)]dx
=x^3 | [-1,0] + acrtanx | [-1,0]
=0^3 - (-1)^3 +0-(-π/4)
=1+π/4
∫[-1,0][(3x^4+3x^2+1)/(1+x^2)]dx
=∫[-1,0] 3x^2dx +∫[-1,0] 1/(1+x^2)]dx
=x^3 | [-1,0] + acrtanx | [-1,0]
=0^3 - (-1)^3 +0-(-π/4)
=1+π/4