The integral I =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits (1)∕(ax + b)\kern 1.66702pt dx, can be done by substitution, z = ax + b, dx = dz∕a, I = {1\over a}\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {1\over z}\kern 1.66702pt dz = {1\over a}(\mathop{ln}\nolimits z + C). Thus
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\class{boxed}{ I =\mathop{ \mathop{\mathop{∫
}\nolimits }}\nolimits (1)∕(ax + b)\kern 1.66702pt dx = {1\over
a}\left (\mathop{ln}\nolimits (ax + b) + C\right . }
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