∫(1到正无穷大)dx/{x^2(x+1)}
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∫(1到正无穷大)dx/{x^2(x+1)}
let
1/[x^2(x+1)] = A/x^2+B/x+C/(x+1)
=> 1=A(x+1) +Bx(x+1)+Cx^2
put x=-1
C = 1
coef. of x^2
B +C =0
B= -1
coef. of constant
A=1
1/[x^2(x+1)] = 1/x^2-1/x+1/(x+1)
∫(1到正无穷大)dx/{x^2(x+1)}
=∫(1到正无穷大)[ 1/x^2-1/x+1/(x+1)] dx
= [-1/x -lnx + ln(x+1)] (1到正无穷大)
= 1-ln2
1/[x^2(x+1)] = A/x^2+B/x+C/(x+1)
=> 1=A(x+1) +Bx(x+1)+Cx^2
put x=-1
C = 1
coef. of x^2
B +C =0
B= -1
coef. of constant
A=1
1/[x^2(x+1)] = 1/x^2-1/x+1/(x+1)
∫(1到正无穷大)dx/{x^2(x+1)}
=∫(1到正无穷大)[ 1/x^2-1/x+1/(x+1)] dx
= [-1/x -lnx + ln(x+1)] (1到正无穷大)
= 1-ln2
∫(1到正无穷大)dx/{x^2(x+1)}
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