Question

求广义积分∫1/x²（x+1）dx 积分区间为【1,

Answer 1

1/x^2(x+1)=(Ax+B)/x^2+C/(x+1)

=[(Ax+B)(x+1)+Cx^2]/x^2(x+1)

=[Ax^2+Ax+Bx+B+Cx^2]/x^2(x+1)

=[(A+C)x^2+(A+B)x+B]/x^2(x+1)

A+C=0

A+B=0

B=1

A=-1

C=1

所以

1/x^2(x+1)=(-x+1)/x^2+1/(x+1)

∫1/x²（x+1）dx

=∫(-x+1)dx/x^2+∫dx/(x+1)

=∫-dx/x+∫dx/x^2+∫d(x+1)/(x+1)

=-lnx+∫dx*x^(-2)+ln(x+1)+C

=ln(x+1)-lnx-1/x+C

=ln(x+1)/x-1/x+C

当x->∞时

原式=ln(x+1)/x-1/x+C=ln(1+1/x)-1/x+C=ln1-0+C=C

当x->1时

原式=ln2-1+C

所以∫1/x²（x+1）dx 积分区间为【1,正无穷）=C-（ln2-1+C）=1-ln2