How do you evaluate the integral #int (x+5)/(3x-1)#?

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Answer 1 · 2017-02-20T13:49:43

The answer is #=1/3x+16/9ln(|3x+1|)+C#

We need

#intdx/x=lnx+C#

Let's rewrite

#x+5=1/3(3x-1)+16/3#

So,

#((x+5))/(3x+1)=1/3*cancel(3x-1)/cancel(3x-1)+16/3*1/(3x+1)#

#int((x+5)dx)/(3x+1)=int1/3dx+16/3intdx/(3x+1)#

#=1/3x+16/3ln(|3x+1|)/3+C#

#=1/3x+16/9ln(|3x+1|)+C#