Question

{(1-x^2/1+x^2)^1/2}*x*dx这个不定积分咋算啊?

Answer 1

令:(1-x^2)/(1+x^2)^1/2 =t ; x^2= (1-t^2)/(1+t^2)

∫(1-x^2)/(1+x^2)^1/2 *x*dx

=1/2∫(1-x^2)/(1+x^2)^1/2 *dx^2

=1/2∫t d[(1-t^2)/(1+t^2)]

=1/2*t*(1-t^2)/(1+t^2) - 1/2∫(1-t^2)/(1+t^2)dt

=1/2*t*(1-t^2)/(1+t^2) - 1/2∫(2-(1+t^2))/(1+t^2)dt

=1/2*t*(1-t^2)/(1+t^2) - ∫1/(1+t^2)dt +1/2∫1dt

=1/2*t*(1-t^2)/(1+t^2) - arctant+1/2t +C

=1/2*(1-x^2)/(1+x^2)^1/2*(1-t^2)/(1+t^2)-arctan(1-x^2)/(1+x^2)^1/2 +1/2(1-x^2)/(1+x^2)^1/2 +C