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

Answer 1

Scarlett Allison

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#

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Answer 2

Isabella Ackerman

To evaluate the integral (\int \frac{{x+5}}{{3x-1}} , dx), we can use the method of partial fraction decomposition. First, we decompose the rational function into simpler fractions, and then integrate each term separately. Once integrated, we combine the results to find the final solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.