高数题,

来源:学生作业帮助网 编辑:作业帮 时间:2024/03/29 01:31:16
高数题,

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).