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Maurinko [17]
3 years ago
15

Integrate: log(ax+b)/(ax+b) dx​

Mathematics
1 answer:
Elza [17]3 years ago
6 0

Answer:

\frac{1}{2a}  \bigg[log(ax + b) \bigg]^{2}  + c

Step-by-step explanation:

\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}}

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