本文主要內容，∫xdx/(ax+b)^2=(1/a^2)[b/(ax+b)+ln|ax+b|]+C的結論證明及其應用。
一、結論的證明
∫xdx/(ax+b)^2
=(1/a)∫xd(ax+b)/(ax+b)^2
=(-1/a)∫xd[1/(ax+b)]
=(-1/a^2)[(ax+b-b)]/(ax+b)]+(1/a)∫dx/(ax+b)
=(-1/a^2)+(1/a^2)*b/(ax+b)+(1/a^2)∫d(ax+b)/(ax+b)
=(1/a^2)*b/(ax+b)+(1/a^2)ln|ax+b|+C.
=(1/a^2)[b/(ax+b)+ln|ax+b|]+C
二、結論的應用舉例
1.當a=1，b=2時：
∫xdx/(x+2)^2
=∫xd(x+2)/(x+2)^2
=-∫xd[1/(x+2)]
=-[(x+2-2)]/(x+2)]+∫dx/(x+2)
=-1+2/(x+2)+∫d(x+2)/(x+2)
=2/(x+2)+ln|x+2|+C.
2.當a=1，b=3時：
∫xdx/(x+3)^2
=∫xd(x+3)/(x+3)^2
=-∫xd[1/(x+3)]
=-[(x+3-3)]/(x+3)]+∫dx/(x+3)
=-1+3/(x+3)+∫d(x+3)/(x+3)
=3/(x+3)+ln|x+3|+C.
3.當a=2，b=1時：
∫xdx/(2x+1)^2
=(1/2)∫xd(2x+1)/(2x+1)^2
=(-1/2)∫xd[1/(2x+1)]
=(-1/4)[(2x+1-1)]/(2x+1)]+(1/2)∫dx/(2x+1)
=(-1/4)+(1/4)/(2x+1)+(1/4)∫d(2x+1)/(2x+1)
=(1/4)/(2x+1)+(1/4)ln|2x+1|+C.
=(1/4)[1/(2x+1)+ln|2x+1|]+C
4.當a=2，b=3時：
∫xdx/(2x+3)^2
=(1/2)∫xd(2x+3)/(2x+3)^2
=(-1/2)∫xd[1/(2x+3)]
=(-1/4)[(2x+3-3)]/(2x+3)]+(1/2)∫dx/(2x+3)
=(-1/4)+(1/4)*3/(2x+3)+(1/4)∫d(2x+3)/(2x+3)
=(1/4)*3/(2x+3)+(1/4)ln|2x+3|+C.
=(1/4)[3/(2x+3)+ln|2x+3|]+C
5.當a=√2，b=1時：
∫xdx/(√2x+1)^2
=(1/√2)∫xd(√2x+1)/(√2x+1)^2
=(-1/√2)∫xd[1/(√2x+1)]
=(-1/2)[(√2x+1-1)]/(√2x+1)]+(1/√2)∫dx/(√2x+1)
=(-1/2)+(1/2)/(√2x+1)+(1/2)∫d(√2x+1)/(√2x+1)
=(1/2)/(√2x+1)+(1/2)ln|√2x+1|+C.
=(1/2)[1/(√2x+1)+ln|√2x+1|]+C
6.當a=3，b=√5時：
∫xdx/(3x+√5)^2
=(1/3)∫xd(3x+√5)/(3x+√5)^2
=(-1/3)∫xd[1/(3x+√5)]
=(-1/9)[(3x+√5-√5)]/(3x+√5)]+(1/3)∫dx/(3x+√5)
=(-1/9)+(1/9)*√5/(3x+√5)+(1/9)∫d(3x+√5)/(3x+√5)
=(1/9)*√5/(3x+√5)+(1/9)ln|3x+√5|+C.
=(1/9)[√5/(3x+√5)+ln|3x+√5|]+C
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