Question

【求不定积分】

前两道用分部积分法
后两道用第二类换元法做
请问具体应该怎么做

Answer 1

1、
∫x^3*e^(-x^2)dx
=∫(1/2)*x^2*e^(-x^2)d(x^2)_________x^2=t
=0.5∫t*e^(-t)dt
=0.5∫(-t)d(e^(-t))
=0.5[(-t)*e^(-t)-∫e^(-t)d(-t)]
=-0.5[t*e^(-t)+e^(-t)]
=-0.5(x^2+1)e^(-x^2)

2、
∫(arcsinx)^2dx__________arcsinx=t
=∫t^2d(sint)
=t^2*sint-∫sintd(t^2)
=t^2*sint-∫2t*sintdt
=t^2*sint+∫2td(cost)
=t^2*sint+2t*cost-∫costd(2t)
=t^2*sint+2t*cost-2∫costdt
=t^2*sint+2t*cost-2sint
=(arcsinx)^2*x+2arcsinx*根号(x^2-1)-2x

3、
∫√(1-2x-x^2) dx
=∫√(2-(x-1)^2) d(x-1)___________x-1=t
=∫√[(√2)^2-t^2] dt___________t=√2*sinu
=∫(√2*cosu) d(√2*sinu)
=∫2cosudsinu
=∫2(cosu)^2du
=∫(1+cos(2u))du
=∫du+0.5∫cos(2u)d(2u)
=u+0.5sin(2u)
=arcsin(t/√2)+0.5*t*√(2-t^2)=。。。偷懒

4、
∫1/√(1+2x^2)dx______________x=(1/√2)*tgt
=∫1/√(1+(tgt)^2)d((1/√2)*tgt)
=(1/√2)*∫1/sect d(tgt)
=(1/√2)*∫cost*(sect)^2d(t)
=(1/√2)*∫sectd(t)__________有公式，算了，还是写一点
=(1/√2)*∫cost/(1-(sint)^2)d(t)
=(1/√2)*∫1/(1-(sint)^2)d(sint)
=(1/√2)*∫1/(1-(sint)^2)d(sint)
=(1/√2)*∫1/(1-sint)d(sint)+∫1/(1+sint)d(sint)...不想写了，看书

Answer 2

楼上没回答完整！

楼主看图！