求定积分，如图

令t^2=1+4x^2，tdt=4xdx
原式=x^3*tdx=x^2*t*xdx=(t^2-1)/4*t*tdt/4=(t^2-1)*t^2/16*dt
=(t^4-t^2)/16*dt
=(t^5/5-t^3/3)/16+C，t的积分区间1到sqrt(5)
计算机计算结果
int(x^3*sqrt(1+4*x^2),0,1) = 5/24*5^(1/2)+1/120
int((t^4-t^2)/16,1,sqrt(5)) = 5/24*5^(1/2)+1/120

∫(0->1) x^3.√(1+4x^2) dx
=(1/12) ∫(0->1) x^2. d(1+4x^2)^(3/2)
=(1/12) [x^2. (1+4x^2)^(3/2)]|(0->1) -(1/6) ∫(0->1) x(1+4x^2)^(3/2) dx
=(1/12).5^(3/2) -(1/120)[ (1+4x^2)^(5/2)]|(0->1)
=(1/12).5^(3/2) -(1/120)5^(5/2) +1/120
=(5/12)√5 - (5/24)√5 +1/120
= (5/24)√5 +1/120

原积分＝1/2 ∫(0, 1)x^2(1+4x^2)^(1/2)d(x^2)
＝1/2 1/16 ∫(0, 1)[(1+4x^2)-1](1+4x^2)^(1/2)d(1+4x^2)
＝1/32 ∫(0, 1)[(1+4x^2)^(3/2)-(1+4x^2)^(1/2)]d(1+4x^2)
＝1/32 [2/5 (1+4x^2)^(5/2)-2/3 (1+4x^(3/2)]|(0, 1)
＝1/32{2/5 * [5^(5/2)-1]-2/3 * [5^(3/2)-1]
＝5√5(1/16-1/48)-1/80+1/48
＝5√5/24-1/120.

希望有所帮助