Question

求解不定积分要过程 ∫arctan√x dx 分部积分法 ∫x²／√(1-x²﹚dx 第二换元法

Answer 1

∫arctan√x dx
令√x=t x=t² dx=2tdt
原式= ∫(arctant)*2tdt
= ∫arctantdt²
=t²arctant-∫t²darctant=t²arctant-∫t²/(1+t²)dt
=t²arctant-∫[1-1/(1+t²)]dt
=t²arctant-t+arctant+C
=xarctan√x-√x+arctan√x+C
∫x²／√(1-x²﹚dx
令x=sint dx=costdt t=arcsinx sin2t=2sintcost=2x√(1-x²)
原式=∫[sin²t/cost]costdt=∫sin²tdt=∫(1-cos2t)/2dt=t/2-sin2t/4=(1/2)arcsinx-(1/2)x√(1-x²)+C

Answer 2

∫arctan√x dx
=xarctan√x-∫xd(arctan√x)
=xarctan√x-∫x*1/(1+x)dx
=xarctan√x-∫(1-1/(1+x))dx
=xarctan√x-x+ln(1+x)+C
∫x²/√(1-x²﹚dx x=sint 0<t<π/2 dx=costdt
=∫ sin^2t/cost *cost dt
=∫ sin^2tdt
=∫ (1-cos2t)/2dt
=t/2-1/4sin2t+C
=1/2arcsinx-1/2x√1-x^2+C