① ∫(2x+4)／(x2 +2x+3) dx； ② ∫（x2）／(1+x2)arctanx dx; ③ 1／[（3√x）+1] dx 注：
问题描述：
① ∫(2x+4)／(x2 +2x+3) dx； ② ∫（x2）／(1+x2)arctanx dx; ③ 1／[（3√x）+1] dx 注：
注：x后的2为平方,根号前的3为开立方；
答
1、原式=∫d(x^2+2x+3)/(x^2+2x+3)+2∫dx/(x^2+2x+3)
=ln|x^2+2x+3|+2∫dx/[(x+1)^2+2]
=ln|x^2+2x+3|+√2∫d[(x+1)/√2]/{[（x+1)/√2]^2+1}
=ln|x^2+2x+3|+√2arctan(x+1)/√2+C.
2、原式=∫（1+x^2-1)arctanxdx/(1+x^2)
=∫arctanxdx-∫arctanxdx/(1+x^2)
=x*arctanx-∫xdx/(1+x^2)-∫arctanxd(arctanx)
=x*arctanx-(1/2)∫d(1+x^2)/(1+x^2)-(1/2)(arctanx)^2
=xarctanx-(1/2)ln(1+x^2)-(1/2)(arctanx)^2+C.
3、设x^(1/3)=t,
x=t^3,
dx=3t^2dt,
原式=∫3t^2dt/(t+1)
=3∫（t+1)dt-6∫dt+3∫d(t+1)/(t+1)
=3(t^2/2+t)-6t+3ln|1+t|+C
=3[x^(2/3)/2+x^(1/3)]-6x^(1/3)+3ln|1+x^(1/3)|+C
=(3/2)x^(2/3) -3x^(1/3)+3ln|1+x^(1/3)|+C.看不懂诶……