y=1/x 与y=x的交点是(1,1)
∴
∫(1->2)dx ∫(1/x ->x) x/(1+y) dy
=∫(1->2)xdx ∫(1/x ->x) dy/(1+y)
=∫(1->2)xdx ∫(1/x ->x) d(y+1)/(1+y)
=∫(1->2)xdx ln(y+1)|(1/x ->x)
=∫(1->2)xdx [ln(x+1) -ln(1/x +1)]
=∫(1->2)xdx [ln(x+1) -ln((x+1)/x)]
=∫(1->2)xdx [ln(x+1) -ln((x+1) + lnx]
=∫(1->2)xlnxdx
=1/2∫(1->2) lnx dx^2
=1/2 x^2 lnx |(1->2) -1/2∫(1->2) x^2 *1/x *dx
=1/2 (4ln2 -1ln1) -1/2∫(1->2) xdx
=2ln2 -1/4* x^2 |(1->2)
=2ln2 -1/4 (4-1)
=2ln2 -1 +1/4
=2ln2 -3/4
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