学帮网{(1-x^2/1+x^2)^1/2}*x*dx这个不定积分咋算啊?
来源：学生作业学帮网 编辑：学帮网 时间：2024/05/30 09:55:42
{(1-x^2/1+x^2)^1/2}*x*dx这个不定积分咋算啊?
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
看图,这类积分少见吧
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