Question

根号下1+x^2的不定积分

Answer 1

I=∫√(1+x^2)dx=x√(1+x^2)-∫[x^2/√(1+x^2)]dx
=x√(1+x^2)-∫[(1+x^2-1)/√(1+x^2)]dx
=x√(1+x^2)-I+∫[1/√(1+x^2)]dx

对于∫[1/√(1+x^2)]dx，令 x=tant, 则 dx=(sect)^2,
I1=∫[1/√(1+x^2)]dx=∫sectdt=∫[sect(sect+tant)/(sect+tant)]dt
=∫{[(sect)^2+secttant]/(sect+tant)}dt=ln|sect+tant|+2C
=ln|x+√(1+x^2)]|+2C,

于是 2I=x√(1+x^2)+ln|x+√(1+x^2)]|+2C,
则 I=(1/2)x√(1+x^2)+(1/2)ln|x+√(1+x^2)]|+C

Answer 2