高等数学不定积分求解
如图
第一题方法可能是唯一的,其他两题给出我的解法。
1.提示:令x^(1/6)=t
详细解答见附图,如不清晰请点击
1。 令x^(1/6)=t,则:x=t^6,dx=(6t^5)dt 且,x^(1/3)=t^2,x^(1/2)=t^3 原式=∫[t^2/t^6*(t^3+t^2)]*(6t^5)dt =6∫[t^7/t^8*(t+1)]dt =6∫dt/[t*(t+1)] =6*∫[(1/t)-(1/t+1)]dt =6ln|t/(t+1)|+C =6ln|x^(1/6)/[1+x^(1/6)]|+C。
2。 令arccosx=t,则x=cost,dx=d(cost)=-sintdt 原式=∫[(cost)^3*t/sint]*(-sint)dt =-∫(cost)^3*tdt =-∫cost*(cos^2 t)*tdt =-∫cost*(1-sin^2 t)*tdt =∫(sin^2 t-1)*td(sint) =∫t*sin^2 td(sint)-∫td(sint) =(1/3)∫td(sin^3 t)-(tsint-∫sintdt) =(1/3)[t*sin^3 t-∫sin^3 tdt]-(tsint+cost) =(1/3)t*sin^3 t-(1/3)∫sint*(sin^2 t)dt-tsint-cost =(1/3)t*sin^3 t-(1/3)∫sint*(1-cos^2 t)dt-tsint-cost =(1/3)t*sin^3 t+(1/3)∫(1-cos^2 t)d(cost)-tsint-cost =(1/3)t*sin^3 t+(1/3)cost-(1/9)cos^3 t-tsint-cost+C =(1/3)arccosx*(√1-x^2)^3-(2x/3)-(x^3/9)-arccosx*√(1-x^2)+C。
3。
原式=∫(sinxdx)/[sin^2 x*(cosx+2)] =-∫d(cosx)/[(1-cos^2 x)*(cosx+2)] =∫d(cosx)/[(cos^2 x-1)*(cosx+2)] =∫d(cosx)/[(cosx+1)*(cosx-1)*(cosx+2)] 令1/[(t+1)*(t-1)*(t+2)]=A/(t+1)+B/(t-1)+C/(t+2) =[A*(t-1)(t+2)+B(t+1)(t+2)+C(t+1)(t-1)]/[(t+1)(t-1)(t+2)] =[A(t^2+t-2)+B(t^2+3t+2)+C(t^2-1)]/[(t+1)(t-1)(t+2)] =[(A+B+C)t^2+(A+3B)t+(-2A+2B-C)]/[(t+1)(t-1)(t+2)] 所以: A+B+C=0 A+3B=0 -2A+2B-C=1 解得:A=-1/2,B=1/6,C=1/3 所以,原积分=(-1/2)∫dt/(t+1)+(1/6)∫dt/(t-1)+(1/3)∫dt/(t+2) =(-1/2)ln|t+1|+(1/6)ln|t-1|+(1/3)ln|t+2|+C =(-1/2)ln|cosx+1|+(1/6)ln|cosx-1|+(1/3)ln|cosx+2|+C。
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