∫dx/((ax＋b)x)＝?

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∫dx/((ax＋b)x)＝?∫dx/((ax＋b)x)＝?∫dx/((ax＋b)x)＝?∫1/[(ax+b)x]dx=(1/b)∫[(ax+b)-ax]/[(ax+b)x]dx=(1/b)∫1/xd

∫dx/((ax＋b)x)＝?
∫dx/((ax＋b)x)＝?

∫dx/((ax＋b)x)＝?
∫ 1/[(ax + b)x] dx
= (1/b)∫ [(ax + b) - ax]/[(ax + b)x] dx
= (1/b)∫ 1/x dx - (a/b)∫ dx/(ax + b)
= (1/b)∫ 1/x dx - (1/b)∫ d(ax + b)/(ax + b)
= (1/b)ln|x| - (1/b)ln|ax + b| + C
= (1/b)ln|x/(ax + b)| + C

1/[x(ax+b)] = k1/x + k2/(ax+b)
1= k1(ax+b)+k2x
x=0
k1= 1/b
coef. of x
k1a+k2=0
k2= -b/a

∫dx/((ax＋b)x)
=(1/b)∫dx/x- (b/a)∫dx/(ax+b)
=(1/b)ln|x|-(b/a^2) ln|ax+b| + C

高数忘完了，来混经验

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