Question

S 1到正无穷1/X（1+X）^2 dx ,S为积分号

Answer 1

求S(1,正无穷)[1/x(1+x)²]dx

1/[x(1+x)²]=[1/x]-[1/(1+x)]-[1/(1+x)²]

所以

S(0,正无穷)[1/x(1+x)²]dx

=[ln(x)-ln(1+x)+(1/(1+x)](1,正无穷)

=[ln(1/((1/x)+1))+(1/(1+x))](1,正无穷)

=[ln(1/(0+1))+0]-[ln(1/2)+1/2]

=-ln(1/2)-(1/2)

=(ln2)-(1/2)