求不定积分 f[x^(3/2)]/(1+x) dxf[x^(3/2)]/(1+x) dx

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/03 07:23:32
求不定积分 f[x^(3/2)]/(1+x) dxf[x^(3/2)]/(1+x) dx

求不定积分 f[x^(3/2)]/(1+x) dxf[x^(3/2)]/(1+x) dx
求不定积分 f[x^(3/2)]/(1+x) dx
f[x^(3/2)]/(1+x) dx

求不定积分 f[x^(3/2)]/(1+x) dxf[x^(3/2)]/(1+x) dx
答:f是积分号吧?
设t=√x,dx=d(t^2)=2tdt
∫ [ (x^(3/2) ] /(1+x) dx
=∫ (t^3) /(1+t^2) *2t dt
= 2 ∫ (t^4) /(1+t^2) dt
=2 ∫ (t^2+1-1)^2 /(1+t^2) dt
=2 ∫ (1+t^2) -2 +1/(1+t^2) dt
=2 ∫ (t^2-1) +1/(1+t^2) dt
=(2/3)*t^3 -2t +2arctant+C
=(2/3)*x^(3/2)-2√x+2arctan(√x)+C

f[x^(3/2)]/(1+x) dx
=fxx^(1/2)/(1+x)dx
=f(1+x-1)/(1+x) *x^(1/2)dx
=f(1-1/(1+x))x^(1/2)DX
=f(x^(1/2)dx-x^(1/2)/(1+x)dx
由于x>0 可以x=tan^2t
dx=2tantsec^2xtdt
后者=ftant/sec^2t*2...

全部展开

f[x^(3/2)]/(1+x) dx
=fxx^(1/2)/(1+x)dx
=f(1+x-1)/(1+x) *x^(1/2)dx
=f(1-1/(1+x))x^(1/2)DX
=f(x^(1/2)dx-x^(1/2)/(1+x)dx
由于x>0 可以x=tan^2t
dx=2tantsec^2xtdt
后者=ftant/sec^2t*2tantsec^2tdt
=f2tan^2tdt
=f2(sec^2t-1)dt
=2tant-2t
=2x^(1/2)-2arctan[(x)^(1/2)]+c
原式=2/3x^(3/2)-2x^(1/2)+2arctan[(x)^(1/2)]+c

收起