两道不定积分求过程

Question

如题所述

Answer 1

1.分解因式

2.

令x=asint,则dx=acost dt ∫x²/√(a²-x²) dx=∫a²sin²t/(acost)·acostdt=a²∫sin²t dt=a²∫(1-cos2t)/2 dt=a²∫1/2dt-a²∫cos2tdt=a²t/2-1/2·a²sin2t+C=1/2·a²arcsin(x/a)-x·√(a²-x²)+C

Answer 2

1、∫x/(x²-x-2)dx=∫x/[(x+1)(x-2)]dx
=∫(x+1-1)/[(x+1)(x-2)]dx
=∫{1/(x-2)-1/3*[(x+1)-(x-2)]/[(x+1)(x-2)]}dx
=∫{1/(x-2)-1/3*[1/(x-2)-1/(x+1)]}dx
=1/3*∫[2/(x-2)+1/(x+1)]dx
=1/3*[2ln|x-2|+ln|x+1|]+C
2、∫x²dx/√(a²-x²)（a>0） 令x=asinθ，则dx=acosθdθ
=∫a²sin²θ*acosθdθ/(acosθ)
=a²/2*∫2sin²θdθ
=a²/2*∫(1-cos2θ)dθ
=a²/2*(θ-1/2*sin2θ)+C
=a²/2*[arcsin(x/a)-1/2*2*x/a*√(1-x²/a²)]+C
=a²/2*[arcsin(x/a)-x√(a²-x²)/a²]+C
=a²/2*arcsin(x/a)-x√(a²-x²)/2+C