大一微積分 請高手教一下

(2)∫(0~pi/4)√[(1+sx)/(1-sx)]dx.....sx=sinx=∫√[(1+sx)^2/(1-sx)(1+sx)]dx=∫(1+sx)dx/cx.....cs=cosx=∫dx/cx+∫sx*dx/cx=∫cx*dx/cx^2-∫d(cx)/cx=∫d(sx)/(1-sx^2)-ln(cx)=∫d(sx)/2(1+sx)+∫d(sx)/(1-sx)-ln(cos45)+ln(cos0)=[ln(1+sx)-ln(1-sx)]/2-ln(1/√2)=0.5*ln[(1+sx)/(1-sx)]+ln(√2)=0.5{ln[(1+1/√2)/(1-1/√2)]-ln(1/1)}+ln√2=0.5*ln[(√2+1)/(√2-1)]+ln√2=0.5*ln[(√2+1)^2/(√2-1)(√2+1)]+ln(√2)=0.5*2*ln(√2+1)+ln√2=ln[√2(√2+1)]=ln(2+√2).....ans=版主答案
(4)∫(1~27)2dx/[x+x^(1/3)].....y=x^(1/3) => x=y^3=∫(1~3)2*3y^2dy/(y^3+y).....dx=3y^2*dy=∫6ydy/(y^2+1)=3∫d(y^2+1)/(y^2+1)=3*ln(y^2+1)=3*ln(10)-3*ln(2)=3*ln(10/2)=3*ln(5).....ans=版主答案
(5) w5=∫(0~1)dx/[1+x^(3/2)]Let y=√x => y^2=x, 2ydy=dx, x^(3/2)=y^3, a=√3/2, b=2/3w5=∫(0~1)2y*dy/(1+y^3)=2∫ydy/(1+y)(1-y+y^2)=-b∫dy/(1+y)+b∫dy/(y^2-y+1)+b∫ydy/(y^2-y+1)=-b*ln(1+y)+b∫dy/[(y-1/2)^2+a^2]+b∫(y-1/2+1/2)dy/[(y-1/2)^2+a^2]=-b*ln(2)+(b+b/2)∫dy/[(y-1/2)^2+a^2]+b∫(y-1/2)dy/[(y-1/2)^2+a^2]=-b*ln(2)+∫dy/[(y-1/2)^2+a^2]+b/2*∫d[(y-1/2)^2+a^2]dy/[(y-1/2)^2+a^2]=-b*ln(2)+(1/a)*atan[(y-1/2)/a]+(1/3)*ln[(y-1/2)^2+a^2]=-b*ln(2)+(1/a)*atan(1/2a)-(1/a)*atan(-1/2a)++[(1/3)*ln(1/4+a^2)-(1/3)*ln(1/4+a^2)]=-b*ln(2)+(2/a)*atan(1/2a)=(2*2/√3)*atan(2/2√3)-b*ln(2)+0=(4/√3)*atan(1/√3)-(2/3)*ln√2=(4/√3)*pi/6-(2/3)*ln√2=2pi/3√3-(2/3)*ln√2=2√3pi/9-(6/9)*ln√2=2(√3*pi-3*ln√2)/9.....ans

第4,5題都是要換變數,
把根號三次x換成y,
積分和根號的符號我不會打,你自己先算算看,
算不出來再問吧.