题目
根号(1+x平方)的积分怎么解
提问时间:2020-12-01
答案
令x=tanα,则:√(1+x^2)=√[1+(tanα)^2]=1/cosα, dx=[1/(cosα)^2]dα.
sinα=√{(sinα)^2/[(sinα)^2+(cosα)^2]}=√{(tanα)^2/[1+(tanα)^2}
=x/√(1+x^2),
∴原式=∫{(1/cosα)[1/(cosα)^2]}dα
=∫[cosα/(cosα)^4]dα
=∫{1/[1-(sinα)^2]^2}d(sinα).
再令sinα=u,则:
原式=∫[1/(1-u^2)^2]du
=(1/4)∫[(1+u+1-u)^2/(1-u^2)^2]du
=(1/4)∫[(1+u)^2/(1-u^2)^2]du+(1/2)∫[(1-u^2)/(1-u^2)^2]du
+(1/4)∫[(1-u)^2/(1-u^2)^2]du
=(1/4)∫[1/(1-u)^2]du+(1/2)∫[1/(1-u^2)]du+(1/4)∫[1/(1+u)^2]du
=-(1/4)∫[1/(1-u)^2]d(1-u)+(1/4)∫[(1+u+1-u)/(1-u^2)]du
+(1/4)∫[1/(1+u)^2]d(1+u)
=(1/4)[1/(1-u)]-(1/4)[1/(1+u)]+(1/4)∫[1/(1-u)]du
+(1/4)∫[1/(1+u)]du
=(1/4)[1/(1-sinα)]-(1/4)[1/(1+sinα)]
-(1/4)∫[1/(1-u)]d(1-u)+(1/4)∫[1/(1+u)]d(1+u)
=(1/4){1/[1-x/√(1+x^2)]}-(1/4){1/[1+x/√(1+x^2)]}
-(1/4)ln|1-u|+(1/4)ln|1+u|+C
=(1/4)[1+x/√(1+x^2)-1+x/√(1+x^2)]/[1-x^2/(1+x^2)]
+(1/4)ln|1+sinα|-(1/4)ln|1-sinα|+C
=(1/4)[2x/√(1+x^2)]/[(1+x^2-x^2)/(1+x^2)]
+(1/4)ln[|1+x/√(1+x^2)|/|1-x/√(1+x^2)|]+C
=(1/2)x√(1+x^2)+(1/4)ln|[√(1+x^2)+x]/[√(1+x^2)-x]|+C
=(1/2)x√(1+x^2)+(1/4)ln|[√(1+x^2)+x]^2/(1+x^2-x^2)|+C
=(1/2)x√(1+x^2)+(1/2)ln|x+√(1+x^2)|+C
sinα=√{(sinα)^2/[(sinα)^2+(cosα)^2]}=√{(tanα)^2/[1+(tanα)^2}
=x/√(1+x^2),
∴原式=∫{(1/cosα)[1/(cosα)^2]}dα
=∫[cosα/(cosα)^4]dα
=∫{1/[1-(sinα)^2]^2}d(sinα).
再令sinα=u,则:
原式=∫[1/(1-u^2)^2]du
=(1/4)∫[(1+u+1-u)^2/(1-u^2)^2]du
=(1/4)∫[(1+u)^2/(1-u^2)^2]du+(1/2)∫[(1-u^2)/(1-u^2)^2]du
+(1/4)∫[(1-u)^2/(1-u^2)^2]du
=(1/4)∫[1/(1-u)^2]du+(1/2)∫[1/(1-u^2)]du+(1/4)∫[1/(1+u)^2]du
=-(1/4)∫[1/(1-u)^2]d(1-u)+(1/4)∫[(1+u+1-u)/(1-u^2)]du
+(1/4)∫[1/(1+u)^2]d(1+u)
=(1/4)[1/(1-u)]-(1/4)[1/(1+u)]+(1/4)∫[1/(1-u)]du
+(1/4)∫[1/(1+u)]du
=(1/4)[1/(1-sinα)]-(1/4)[1/(1+sinα)]
-(1/4)∫[1/(1-u)]d(1-u)+(1/4)∫[1/(1+u)]d(1+u)
=(1/4){1/[1-x/√(1+x^2)]}-(1/4){1/[1+x/√(1+x^2)]}
-(1/4)ln|1-u|+(1/4)ln|1+u|+C
=(1/4)[1+x/√(1+x^2)-1+x/√(1+x^2)]/[1-x^2/(1+x^2)]
+(1/4)ln|1+sinα|-(1/4)ln|1-sinα|+C
=(1/4)[2x/√(1+x^2)]/[(1+x^2-x^2)/(1+x^2)]
+(1/4)ln[|1+x/√(1+x^2)|/|1-x/√(1+x^2)|]+C
=(1/2)x√(1+x^2)+(1/4)ln|[√(1+x^2)+x]/[√(1+x^2)-x]|+C
=(1/2)x√(1+x^2)+(1/4)ln|[√(1+x^2)+x]^2/(1+x^2-x^2)|+C
=(1/2)x√(1+x^2)+(1/2)ln|x+√(1+x^2)|+C
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