Question

求定积分上限e下限1,xln xdx,上限e-1下限1,ln(1+x)dx？

Answer 1

定积分上限e下限1,xln xdx,
=∫(1,e)lnxd(x^2)/2
x^2/2*lnx|(1,e)-∫(1,e)(x^2)/2dlnx
=e^2/2-x^2/4|(1,e)
=e^2/2-e^2/4+1/4
=1/4*(e^2+1)
上限e-1下限1,ln(1+x)dx
=xln(1+x)|(1,e-1)-∫(1,e-1)xdln(1+x)
=(e-1)-ln2-∫(1,e-1)x/(x+1)dx
=(e-1)-ln2-∫(1,e-1)(x+1-1)/(x+1)dx
=e-1-ln2-x|(1,e-1)+ln(1+x)|(1,e-1)
=e-1-ln2-e+2+1-ln2
=2-2ln2,9,上限e-1下限1,ln(1+x)dx
=xln(1+x)|(1,e-1)-∫(1,e-1)xdln(1+x)
=(e-1)-ln2-∫(1,e-1)x/(x+1)dx
=(e-1)-ln2-∫(1,e-1)(x+1-1)/(x+1)dx
=e-1-ln2-x|(1,e-1)+ln(1+x)|(1,e-1)
=e-1-ln2-e+2+1-ln2
=2-2ln2,0,