# How do you find the definite integral of int 5/(3x+1) from [0,4]?

Aug 6, 2017

$\frac{5}{3} \ln 13$

#### Explanation:

To integrate this, one can use substitution method, letting 3x+1 =u. This gives 3dx= du, or dx = $\frac{1}{3} \mathrm{du}$

For x=0, u would be 1 and for x= 4, u would be 13. The given integral thus becomes

${\int}_{u = 1}^{13} \frac{5}{3} \frac{1}{u} \mathrm{du}$

= $\frac{5}{3} {\left[\ln u\right]}_{1}^{13}$

=$\frac{5}{3} \ln 13$

Aug 6, 2017

$\frac{5}{3} \ln 13$

#### Explanation:

$\text{using the standard integral}$

•color(white)(x)int1/(ax+b)dx=1/aln|ax+b|+c

$\Rightarrow {\int}_{0}^{4} \frac{5}{3 x + 1} \mathrm{dx}$

$= {\left[5 . \frac{1}{3} \ln \left(3 x + 1\right)\right]}_{0}^{4}$

$= \frac{5}{3} \ln 13 - {\cancel{\frac{5}{3} \ln 1}}^{0}$

$= \frac{5}{3} \ln 13$