高等数学大一勾的五道题
展开全部
(2) I = ∫[(lnx)^2/x^2]dx = ∫(lnx)^2d(-1/x)
= -(lnx)^2/x +∫(2lnx/x^2)dx
= -(lnx)^2/x + 2∫lnxd(-1/x)
= -(lnx)^2/x - 2lnx/x +∫(1/x^2)dx
= -(lnx)^2/x - 2lnx/x - 1/x + C
(4) I = (1/2)∫x^2 e^(-x^2)dx^2 = (1/2)∫u e^(-u)du
= -(1/2)∫u de^(x) = -(1/2)ue^(-u) + (1/)∫e^(-u)du
= -(1/2)ue^(-u) - (1/2)e^(-u) + C
= -(1/2)(1+x^2)e^(-x^2) + C
(6) I = xarctanx -∫[x/(1+x^2)]dx
= xarctanx - (1/2)ln(1+x^2)+ C
(8) I = xarctan√x -∫[x/(1+x)]dx
= xarctan√x - x + ln(1+x)+ C
(10) I = x(arcsinx)^2 - 2∫[xarcsinx/√(1-x^2)]dx
= x(arcsinx)^2 + 2∫(arcsinx)d√(1-x^2)
= x(arcsinx)^2 + 2arcsinx√(1-x^2) - 2∫dx
= x(arcsinx)^2 + 2arcsinx√(1-x^2) - 2x + C
= -(lnx)^2/x +∫(2lnx/x^2)dx
= -(lnx)^2/x + 2∫lnxd(-1/x)
= -(lnx)^2/x - 2lnx/x +∫(1/x^2)dx
= -(lnx)^2/x - 2lnx/x - 1/x + C
(4) I = (1/2)∫x^2 e^(-x^2)dx^2 = (1/2)∫u e^(-u)du
= -(1/2)∫u de^(x) = -(1/2)ue^(-u) + (1/)∫e^(-u)du
= -(1/2)ue^(-u) - (1/2)e^(-u) + C
= -(1/2)(1+x^2)e^(-x^2) + C
(6) I = xarctanx -∫[x/(1+x^2)]dx
= xarctanx - (1/2)ln(1+x^2)+ C
(8) I = xarctan√x -∫[x/(1+x)]dx
= xarctan√x - x + ln(1+x)+ C
(10) I = x(arcsinx)^2 - 2∫[xarcsinx/√(1-x^2)]dx
= x(arcsinx)^2 + 2∫(arcsinx)d√(1-x^2)
= x(arcsinx)^2 + 2arcsinx√(1-x^2) - 2∫dx
= x(arcsinx)^2 + 2arcsinx√(1-x^2) - 2x + C
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询