How do you evaluate the definite integral of 0 to 4 for #(5 / 3x +1) dx#?

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To evaluate the definite integral of the function ( \frac{5}{3x + 1} ) from 0 to 4, you first need to find its antiderivative. Then, you evaluate the antiderivative at the upper limit of integration (4) and subtract the value of the antiderivative at the lower limit of integration (0). The definite integral of the function from 0 to 4 is:

[ \int_{0}^{4} \frac{5}{3x + 1} dx = \left[ \frac{5}{3} \ln|3x + 1| \right]_{0}^{4} ]

Now, substitute the upper limit (4) and the lower limit (0) into the antiderivative and subtract:

[ \left[ \frac{5}{3} \ln|3(4) + 1| \right] - \left[ \frac{5}{3} \ln|3(0) + 1| \right] ]

[ = \left[ \frac{5}{3} \ln|13| \right] - \left[ \frac{5}{3} \ln|1| \right] ]

Since the natural logarithm of 1 is 0:

[ = \frac{5}{3} \ln|13| - 0 ]

[ = \frac{5}{3} \ln|13| ]

Therefore, the value of the definite integral of ( \frac{5}{3x + 1} ) from 0 to 4 is ( \frac{5}{3} \ln|13| ).