当x∈[0,1] 求定积分∫x^(3⼀2)⼀(1+x) dx

2024-12-03 02:27:25
推荐回答(1个)
回答1:

令(x)^(1/2)=t

∫x^(3/2)/(1+x) dx
=∫2t^4dt/(1+t^2)
=∫[2(t^2+1)^2-4(t^2+1)+2]dt/(1+t^2)
=∫2(t^2+1)-4+2/(1+t^2)dt
=2t^3/3-2t+2arctant+C

故定积分的值为 pi/2-4/3