Answer:
[tex] \frac{1}{2a} \bigg[log(ax + b) \bigg]^{2} + c[/tex]
Step-by-step explanation:
[tex] \int \frac{log(ax + b)}{(ax + b)} dx \\ \\ let \: log(ax + b) = t \\ \\ \frac{1}{(ax + b)} .a \: dx = dt \\ \\ \frac{1}{(ax + b)} \: dx = \frac{1}{a} dt \\ \\ \therefore \: \int \frac{log(ax + b)}{(ax + b)} dx = \int t. \frac{1}{a} dt \\ \\ = \frac{1}{a} \int t. dt \\ \\ = \frac{1}{a} \times \frac{ {t}^{2} }{2} + c \\ \\ = \frac{1}{2a} \bigg[log(ax + b) \bigg]^{2} + c \\ \\ \purple{ \bold{\therefore \: \int \frac{log(ax + b)}{(ax + b)} dx = \frac{1}{2a} \bigg[log(ax + b) \bigg]^{2} + c}}[/tex]
