f(x)=1/π【(1-x^2)^(-1/2)】+xln(x+1)∫(上1下0)f(x)dx

设∫(0,1)f(x)dx=a
f(x)=1/π【(1-x^2)^(-1/2)】+xln(x+1)a,于是
a=∫(0,1)f(x)dx=1/π∫(0,1)【(1-x^2)^(-1/2)】+a∫(0,1)xln(x+1)
=1/πarcsinx|(0,1)+a/2∫(0,1)ln(x+1)dx^2
=1/2+a/2*x^2ln(x+1)∫(0,1)-(a/2)∫(0,1)x^2/(x+1)dx
=1/2+a/2*ln(2)-(a/2)∫(0,1)(x-1+1/(x+1))dx
=1/2+a/2*ln(2)-(a/2)(x^2/2-x+ln(x+1))|(0,1)
=1/2+a/2*ln(2)-(a/2)(1/2-1+ln(2))
=1/2+a/4
a=2/3
所以：f(x)=1/π【(1-x^2)^(-1/2)】+(2/3)xln(x+1)
不知计算错没有,方法绝对正确