# Thread: Am I Wrong (part 2)?

1. ## Am I Wrong (part 2)?

Problem 15: $\int \sqrt{12 + 4x^2}$

I get:

$x\sqrt{x^2 + 3}+ 3 \ln(\frac{\sqrt{x^2 + 3} + x}{\sqrt{3}})$

the book has the same answer except the radical in the denominator of the natural log part is gone. Where did I go wrong?

Thanks

2. What substitution did you make?

I would choose $\displaystyle x = \sqrt{3}\tan{u}$

$\displaystyle \int \sqrt{12+4x^2}~dx = \sqrt{x^2+3}+3\sinh^{-1}\left(\frac{x}{\sqrt{3}}\right)+C$

3. Thats an indefinite integral so your answer should be $x\sqrt{x^2+ 3}+ 3ln(\frac{\sqrt{x^2+ 3}+ x}{\sqrt{3}})+ C$ which is the same as $x\sqrt{x^2+ 3}+ 3ln(\sqrt{x^2+ 3}+ x)- 3ln(\sqrt{3})+ C$ which is the same as $x\sqrt{x^2+ 3}+ 3ln(\sqrt{x^2+ 3}+ x)+ C'$ with $C'= C+ 3ln(\sqrt{3})$.

4. Originally Posted by HallsofIvy
Thats an indefinite integral so your answer should be $x\sqrt{x^2+ 3}+ 3ln(\frac{\sqrt{x^2+ 3}+ x}{\sqrt{3}})+ C$ which is the same as $x\sqrt{x^2+ 3}+ 3ln(\sqrt{x^2+ 3}+ x)- 3ln(\sqrt{3})+ C$ which is the same as $x\sqrt{x^2+ 3}+ 3ln(\sqrt{x^2+ 3}+ x)+ C'$ with $C'= C+ 3ln(\sqrt{3})$.
Right you are sir. Thank you