1. Trig Substitution Verification

I have just worked the following integral and I believe my text has a wrong answer. If anyone would take the time to verify, I would appreciate it.

Initial Problem: $\int\sqrt{4 + 9x^2} dx$

My solution: $\frac{1}{8}[3x\sqrt{4+9x^2} + 4ln|{\sqrt{4+9x^2}+3x}|] + C$

Book solution: $\frac{1}{6}[3x\sqrt{4+9x^2} + 4ln|{\sqrt{4+9x^2}+3x}|] + C$

As you can see, the solutions differ only by the factor's 1/8 and 1/6.
I would appreciate someone verifying for me.

2. please find derivative, to prove, what is correct

3. If you gave us working it would be easier but Maple gives...

$\frac{1}{2}x\sqrt{4+9x^2} + \frac{2}{3} \textrm{arcsinh} \left(\frac{3x}{2} \right)$

which would seem the book is right...

4. Thanks. That's what I needed to know. My work was rather lengthy and on 2 pages, so I didn't include it. Thanks again.

5. Back to the drawing board!

6. Originally Posted by kaiser0792
Back to the drawing board!
Just go over your working again, it's not a massive error so you likely just wrote down the wrong number or something... If you can't find the error post up what substitution you used and if I stop raging at my uni work I might have a look at it.

7. Z= $\int\sqrt{4+9x^2}dx$= $x\sqrt{4+9x^2}$- $\int{x}d(\sqrt{4+9x^2})$=
=(1)- $\int\frac{9x^2dx}{\sqrt{4+9x^2}}$=(1)- $\int\frac{9x^2+4-4}{\sqrt{4+9x^2}}dx$=
=(1)-Z+4 $\int\frac{dx}{\sqrt{4+9x^2}}$
then we get
2Z= $x\sqrt{4+9x^2}$+4 $\int\frac{dx}{\sqrt{4+9x^2}}$
the last int is standart.