Everyone's favorite: Trigonometric substitution to solve integrals help

The question is
∫ (dx) / [ √(9x^2 + 4) ].
If you know anything about this, I will spare myself some time and just skip to my problem...

I end with the answer
(1/3) ln [ |√(9x^2 + 4) +3x| / 2 ] + C

Apparently this is NOT correct because you have to use the /2 part as a constant. So they correct answer is actually

(1/3)ln | √(9x^2 + 4) +3x | + C


I feel like taking the the /2 out is a little bit of a sketch maneuver. How am I supposed to remember this?
Thanks

your answer IS correct... just not completely simplified (by most explanations for "simplified").


I'd ask your instructor if he/she feels it is important at this stage. If he/she does feel it important, then just try to recall the (main 3) properties for expanding logarithms.

I tell my students they "should" remember it, but it is not absolutely necessary. I would certainly accept the result you provided.

this type of result is also argued about among teachers/professors.

even textbook authors can't agree.


It is important (in my opinion) that you realize the two results represent the same thing without having to have it explained. However, unless you are in my class, I would suggest you simply ask your instructor/professor/teacher.

good luck.

First one is the correct answer.

∫ (dx) / [ √(9x^2 + 4) ]

let 3x = 2 tan(u) ==> tan(u) = (3/2)x, sec(u) = √(1 + tan^2(u)) = 1/2√(9x^2 + 4)
3 dx = 2 sec^2(u) du
dx = (2/3)sec^2(u) du

∫(2/3) sec^2(u) du / [ √(4tan^2(u) + 4) ]

= ∫(2/3) sec^2(u) du /2√(1 + tan^2(u))

= 1/3∫sec^2(u) du / sec(u)

= 1/3∫sec (u) du

= 1/3 ln I sec u + tan u I + C

substitute tan(u) = (3/2)x and sec(u) = 1/2√(9x^2 + 4)

= 1/3 ln I 1/2[√(9x^2 + 4) + 3x ] I + C