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

∫sinaxcosbxdx(a,b不等于0)

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)