What is the Correct Integration by Substitution for \int \frac{3x}{2x+3}?

[tweety1234]

Homework Statement

[tex] \int \frac{3x}{2x+3} [/tex]

[tex] u = 2x +3 [/tex]

[tex] x = \frac{1}{2}(u-3} )[/tex]

[tex] dx = \frac{1}{2} du [/tex]

so now the integral should be,

[tex] \int \frac{ \frac{3u-9}{2}}{u} \times \frac{1}{2} du [/tex]

= [tex] \frac{1}{2} \int \frac{3u-9}{2} \times \frac{1}{u} du [/tex]

[tex] \frac{1}{2} \int \frac{3u-9}{2u} du [/tex]

[tex] \frac{1}{2} \int \frac{3u}{2u} - \frac{9}{2u} [/tex]

[tex] \frac{1}{2} \int \frac{3}{2} - \frac{9}{2u} [/tex]

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

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

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

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

thanks

 

[gabbagabbahey]

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

 

[tweety1234]

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

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

so both answers are valid?

 

[Tedjn]

Yes, so the difference between the two is a constant equal to 9/4 + ln(2).