作业帮高数题,
来源：学生作业帮助网 编辑：作业帮 时间：2024/07/17 07:36:44
xTn@(a"4Ɛ&.¢ꊗTAe7| ?*RqWBc{{ft{?!X{ʶ/ooU& AQG㪫Ua_u Vy!O$*G^He:`NgP(1QƩK1B1آä RFZ(kpjA12ecB ńy7^z yRٗM >:7rF9Bڗ .<89:(#P6=g`%o6hpMRS(64iށзrPsr=aMء 7O?kek2%9IG֗hC"oYt)ܢHf @,u
x
高数题,
高数题,
高数题,
原式=x→0lim[-x/√(4-x²)]/(2sin2x)=x→0lim[-x/√(4-x²)]/(4x)=x→0lim{-1/[4√(4-x²)]}=-1/16.
原式=【0,1】∫ln(1+x)d[1/(2-x)]=【0,1】{[ln(1+x)]/(2-x)]-∫dx/(2-x)(1+x)]}
=ln2+【0,1】∫dx/(x-2)(x+1)=ln2+【0,1】∫[1/(x-2)-1/(x+1)]dx
=ln2+[ln∣x-2∣-ln∣x+1∣]【0,1】=ln2+[-ln2-ln2]=-ln2
【0,1】∫dx/√[16-(x-3)²]=【0,1】∫d(x-3)/√[4²-(x-3)²]=arcsin[(x-3)/4]【0,1】
=arcsin(-1/2)-arcsin(-3/4)=-π/6+arcsin(3/4)
原式=【0,1】{xln[x+√(1+x²)]-∫xdx/√(1+x²)}=ln(1+√2)-【0,1】(1/2)∫d(1+x²)/√(1+x²)
=ln(1+√2)-[√(1+x²)]【0,1】=ln(1+√2)-(√2-1)=1-√2+ln(1+√2).
