How do you evaluate the integral #intdx/(2x+3)#?

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Answer 1 · 2017-02-07T19:41:57

The answer is #==1/2ln(|2x+3|)+C#

We need

#intdx/x=lnx+C#

Let #u=2x+3#

#du=2dx#

#dx=du(1/2)#

#intdx/(2x+3)=1/2int(du)/u#

#=1/2lnu#

#=1/2ln(|2x+3|)+C#