作业帮∫xln(x+√(1+x^2))dx
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/03 01:29:22
∫xln(x+√(1+x^2))dx
∫xln(x+√(1+x^2))dx
∫xln(x+√(1+x^2))dx
∫xln(x+√(1+x^2))dx
=1/2∫ln(x+√(1+x^2))dx^2
=1/2ln(x+√(1+x^2))·x^2-1/2∫x^2dln(x+√(1+x^2))
=1/2*x^2*ln(x+√(1+x^2))-1/2∫x^2/√(1+x^2) dx
=1/2*x^2*ln(x+√(1+x^2))-1/2∫tan^2t/sect *sec^2tdt (x=tant)
=1/2*x^2*ln(x+√(1+x^2))-1/2∫tan^2tsectdt
=1/2*x^2*ln(x+√(1+x^2))-1/2∫(sec^2t-1)sectdt
=1/2*x^2*ln(x+√(1+x^2))-1/2∫(sec^3t-sect)dt
因为∫(sec^3t)dt=∫(sect)dtant=secttant-∫tan^2tsectdt
=secttant-∫(sec^3t-sect)dt
2∫sec^3tdt=secttant+∫sectdt
∫sec^3tdt=1/2[secttant+∫sectdt]
所以
原式==1/2*x^2*ln(x+√(1+x^2))-1/4[secttant+∫sectdt]+1/2∫sectdt
=1/2*x^2*ln(x+√(1+x^2))-1/4secttant+1/4∫sectdt\
=1/2*x^2*ln(x+√(1+x^2))-1/4 x√(1+x^2)+1/4ln|sect+tant|+c
=1/2*x^2*ln(x+√(1+x^2))-1/4 x√(1+x^2)+1/4ln|√1+x^2+x|+c
应该学过分部积分法吧~
∫xln(x+√(1+x^2))dx
=1/2∫ln(x+√(1+x^2))dx^2
=1/2*x^2*ln(x+√(1+x^2))-1/2∫x^2dln(x+√(1+x^2))
=1/2*x^2*ln(x+√(1+x^2))-1/2∫x^2*1/(x+√(1+x^2))*[1+x/√(1+x^2)]dx
=1/2*x^2*ln(x+...
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应该学过分部积分法吧~
∫xln(x+√(1+x^2))dx
=1/2∫ln(x+√(1+x^2))dx^2
=1/2*x^2*ln(x+√(1+x^2))-1/2∫x^2dln(x+√(1+x^2))
=1/2*x^2*ln(x+√(1+x^2))-1/2∫x^2*1/(x+√(1+x^2))*[1+x/√(1+x^2)]dx
=1/2*x^2*ln(x+√(1+x^2))-1/2∫x^2/√(1+x^2)dx
=1/2*x^2*ln(x+√(1+x^2))-1/2∫(x^2+1-1)/√(1+x^2)dx
=1/2*x^2*ln(x+√(1+x^2))-1/2∫[√(x^2+1)-1/√(1+x^2)]dx
=1/2*x^2*ln(x+√(1+x^2))-1/2*[x/2*√(x^2+1)+a^2/2*ln|x+√(x^2+1)|]+1/2*ln|x+√(x^2+1)|+C
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