求不定积分：积分号(x的立方⼀(1加x平方)的3⼀2次方)dx

2024-11-08 09:05:14

推荐回答（4个）

回答（1）：

x=tant,t=arctanx,dx=(sect)^2dt
积分号(x的立方/(1加x平方)的3/2次方)dx
=S((tant)^3/(sect)^3*)(sect)^2dt
=S(tant)^3/sect dt
=S(sint)^3/(cost)^2dt
=-S(sint)^2/(cost)^2dcost
=-S(1-(cost)^2)/(cosx)^2dcost
=-S(cost)^(-2)dcost+Sdcost
=1/cost+cost+c
=根号（x^2+1)+1/根号（x^2+1)+c

回答（2）：

解：∫x³dx/(1+x²)^(3/2)=(1/2)∫x²d(1+x²)/(1+x²)^(3/2)
=(1/2)∫[(1+x²-1)/(1+x²)^(3/2)]d(1+x²)
=(1/2)∫[1/(1+x²)^(1/2)-1/(1+x²)^(3/2)]d(1+x²)
=(1/2)[2√(1+x²)+2/√(1+x²)]+C (C是积分常数)
=√(1+x²)+1/√(1+x²)+C。

回答（3）：

x = tant, (1+x²)^(3/2) = sec³t ， dx = sec²t dt
I = ∫tan³t cost dt = ∫tan²t d(-cost)
= - tan²t cost + ∫2 sect tant dt
= - tan²t cost + 2 sect + C
= - x²/ (1+x²)^(1/2) + 2(1+x²)^(1/2) + C

回答（4）：

∫x^3dx/(1+x^2)^(3/2)
=(1/2)∫x^2dx^2/(1+x^2)^(3/2)
=(1/2)∫d(x^2+1)/(1+x^2)^(1/2) -(1/2)∫d(x^2+1)/(1+x^2)^(3/2)
=(1+x^2)^(1/2) +(1+x^2)^(-1/2) + C