∫(x^2+1)/(x^4+1)dx有理数求积分?

Question

∫(x^2+1)/(x^4+1)dx有理数求积分?

Answer 1

被积函数分子分母除以x²有∫(x^2+1)/(x^4+1)dx = ∫(1+1/x²)/(x²+1/x²)dx令u=x-1/x , 则 du = (1+1/x²)dx且 u² = x²+1/x² -2则原式= ∫ du/(u²+2)=1/根号2 * arctan (...