Question

帮我解答一道不定积分题书上的过程看不懂,希望哪位大侠能写下纤细的过程供参考∫2x^2-x/x+1dx∫sinaxcosb

Answer 1

1.(2x^2-x)/(x+1)=2x-3+3/(x+1)

∫2x^2-x/x+1dx =∫[2x-3+3/(x+1)]dx

=x^2-3x+3ln|x+1|+C,C为任意实数；

2.积化和差

sin(a+b)x=sinaxcosbx+cosaxsinbx

sin(a-b)x=sinaxcosbx-cosaxsinbx

则sinaxcobx=[sin(a+b)x+sin(a-b)x]/2

∫sinaxcosbxdx

=∫[sin(a+b)x+sin(a-b)x]/2dx

=-[cos(a+b)x/(a+b)+cos(a-b)/(a-b)]/2

(a,b不等于0)