Question

求1/根号下(x的平方+1)的三次方的不定积分？

Answer 1

令x=tan t 则：dx=sec^2(t)dt
原式=∫sec^2(t)dt/sec^3(t)=∫cos tdt=sin t+C=sin(arctan x)+C=x/√(x^2+1)+C
验证：[x(x^2+1)^(-1/2)+C]'=(x^2+1)^(-1/2)-1/2*2* x^2 (x^2+1)^(-3/2)
=[x^2+1-x^2](x^2+1)^(-3/2)=1/](x^2+1)^(3/2) 完全正确.,4,
kadray 举报
sin(arctan x)+C=x/√(x^2+1)+C怎么化得 tan(arctan x)=x cot(arctan x)=1/x [csc(arctan x)]^2=1+1/x^2=(x^2+1)/x^2 sin(arctan x)+C=x/√(x^2+1),