# Homework Help: Integration by substitution,

1. Jun 13, 2010

### tweety1234

1. The problem statement, all variables and given/known data

$$\int \frac{3x}{2x+3}$$

$$u = 2x +3$$

$$x = \frac{1}{2}(u-3} )$$

$$dx = \frac{1}{2} du$$

so now the integral should be,

$$\int \frac{ \frac{3u-9}{2}}{u} \times \frac{1}{2} du$$

= $$\frac{1}{2} \int \frac{3u-9}{2} \times \frac{1}{u} du$$

$$\frac{1}{2} \int \frac{3u-9}{2u} du$$

$$\frac{1}{2} \int \frac{3u}{2u} - \frac{9}{2u}$$

$$\frac{1}{2} \int \frac{3}{2} - \frac{9}{2u}$$

= $$\frac{1}{2} [ \frac{3}{2}u - \frac{9}{2} ln(2u)]$$

so final answer, after simplifying is ; $$\frac{3}{2}x +\frac{9}{4} - \frac{9}{4}ln(4x+6) +c$$

but my book says the correct answer is $$\frac{3}{2}x - \frac{9}{4}ln(2x+3) +c$$

So if anyone can tell me where I have gone wrong, I would really appreciate it.

thanks

2. Jun 13, 2010

### gabbagabbahey

The two answers are equivalent, just with different constants of integration. Remember, $\ln(4x+6)=\ln(2)+\ln(2x+3)$, so the additional terms are just constants which can be absorbed into the integration constant.

3. Jun 13, 2010

### tweety1234

Oh okay thanks, I didn't think of it like that, are you also referring to the extra 9/4 term that I get?