Question

已知x属于R，比较（x的平方+1）括号的平方与x的4次方+x的平方+1的大小

Answer 1

∫ 1/(x?+1) dx =(1/2)∫ [(1+x2)+(1-x2)]/(x?+1) dx =(1/2)∫ (1+x2)/(x?+1) dx + (1/2)∫ (1-x2)/(x?+1) dx 分子分母同除以x2 =(1/2)∫ (1/x2+1)/(x2+1/x2) dx + (1/2)∫ (1/x2-1)/(x2+1/x2) dx 将分子放到微分符号后 =(1/2)∫ 1/(x2+1/x2) d(x-1/x) - (1/2)∫ 1/(x2+1/x2) d(x+1/x) =(1/2)∫ 1/(x2+1/x2-2+2) d(x-1/x) - (1/2)∫ 1/(x2+1/x2+2-2) d(x+1/x) =(1/2)∫ 1/[(x-1/x)2+2] d(x-1/x) - (1/2)∫ 1/[(x+1/x)2-2] d(x+1/x) =(√2/4)arctan[(x-1/x)/√2] - (√2/8)ln|(x+1/x-√2)/(x+1/x+√2)| + C

追问

什么乱七八糟的