大一高数不定积分问题

Question

第8题。。。都没思路，能做的大神帮忙做一下吧，拜托了

Answer 1

8.（1）∫x^3dx/(x+3)=∫(x^3+27-27)dx/(x+3)=∫(x^2-3x+9-27/(x+3))dx
=x^3/3-3/2x+9x-27ln|x+3|+C
（2）∫dx/sin^2xcosx
令t=sinx,x=arcsint,dx=dt/√(1-t^2)
原式=∫1/[t^2√(1-t^2)]*dt/√(1-t^2)=∫dt/[t^2(1-t^2)]
=∫[1/t^2+1/(1-t^2)]dt=-1/t+1/2ln|(1+t)/(1-t)|+C=-cscx+1/2ln[(1+sinx)/(1-sinx)]
（3）∫dx/(x^3+1)=∫dx/[(x+1)(x^2-x+1)]=1/3∫[1/(x+1)+(-x+2)/(x^2-x+1)]dx
=1/3ln|x+1|+1/3∫[-1/2(2x-1)+3/2]/(x^2-x+1)dx
=1/3ln|x+1|-1/6ln(x^2-x+1)+1/2∫dx/[(x-1/2)^2+3/4]
=1/3ln|x+1|-1/6ln(x^2-x+1)+2/3∫dx/[(√3/2(x-1/2))^2+1]
=1/3ln|x+1|-1/6ln(x^2-x+1)+4√3/9arctan(√3/2(x-1/2))+C
（4）∫(5x-1)/(x^3-x^2+x-1)dx=∫(5x-1)/[(x^2+1)(x-1)]dx
=∫[(3x-2)/(x^2+1)-3/(x-1)]dx=3/2ln(x^2+1)-2arctanx-3ln|x-1|+C
（5）∫dx/(3+cosx)
令u=tan(x/2),x=2arctanu,dx=2du/(1+u^2),cosx=(1-u^2)/(1+u^2)
原式=∫2du/[3(1+u^2)+(1-u^2)]=∫du/(2+u^2)=√2/2arctan(√2/2u)+C
=√2/2arctan(√2/2tan(x/2))+C
（6）∫dx/(1+sinx)
令u=tan(x/2),x=2arctanu,dx=2du/(1+u^2),sinx=2u/(1+u^2)
原式=∫2du/(1+u^2+2u)=-2/(1+u)+C=-2/(1+tan(x/2))+C