Question

求广义积分∫1/x²（x+1）dx 积分区间为【1，正无穷）在线等

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

Answer 2

∫1/x²（x+1）dx =∫[(-x+1)x²/+1/(x+1)]dx
=∫[(-x+1)/x²]dx+∫[1/(x+1)]dx
=∫(-1/x)dx+∫(1/x²)dx+∫[1/(x+1)]dx
=-lnx-1/x+ln(x+1)+c
=ln[(x+1)/x]-1/x+c
=ln(1+1/x)-1/x+c
所以∫1/x²（x+1）dx 在积分区间为[1，正无穷）的植为(ln1-0+c)-(ln2-1+c)=1-ln2