求∫上限1下限-1（7x^3-5x+1）╱（x^2+1）dx

Question

求∫上限1下限-1 （7x^3-5x+1）╱（x^2+1）dx

Answer 1

∫(-1->1) （7x^3-5x+1）╱（x^2+1）dx
=∫(-1->1) （7x^3+7x-12x+1）╱（x^2+1）dx
=∫(-1->1) [7x -12x/(x^2+1) +1/(x^2+1)]dx
=∫(-1->1) 7xdx-∫(-1->1) 12xdx/(x^2+1)+∫(-1->1) dx/(x^2+1)
∵ 7xdx为奇函数 12xdx/(x^2+1)也为奇函数
∴
原式=0-0+arctanx|(-1->1)
=π/4-(-π/4)
=π/2