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,

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

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

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

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

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