Question

一道高数题求助

求解这道题接下来怎么做

Answer 1

换元x=tant就好了

追答

第二行右端少写了dx，见谅

Answer 2

I = ∫x e^arctanx/(1+x^2)^(3/2) dx = x e^arctanx/√(1+x^2) - I
I = (1/2)x e^arctanx/√(1+x^2) + c

Answer 3

∫[(xe^arctanx)/(1+x²)^(3/2)]dx【令arctanx=u，则x=tanu,dx=sec²udu】
=∫[(tanu)(e^u)/(1+tan²u)^(3/2)]sec²udu=∫[(tanu)(e^u)/secu]du
=∫(sinu)(e^u)du=∫sinud(e^u)=(sinu)(e^u)-∫(e^u)cosudu
=(sinu)(e^u)-∫(cosu)d(e^u)=(sinu)(e^u)-[(cosu)e^u+∫(e^u)sinudu]
=(sinu)(e^u)-(cosu)e^u-∫(e^u)sinudu；
移项得;2∫(sinu)(e^u)du=(sinu-cosu)e^u
∴原式=∫(sinu)(e^u)du=(1/2)(sinu-cosu)e^u+C
=(1/2)[x/√(1+x²)-1/√(1+x²)]e^(arctanx)+C
=(x-1)/[2√(1+x²)]e^(arctanx)+C;