Am I Wrong (part 2)?

**Warrenx** · December 8th 2010, 01:03 PM

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

**pickslides** · December 8th 2010, 01:13 PM

What substitution did you make?

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

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

**HallsofIvy** · December 9th 2010, 02:53 AM

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 $C'= C+ 3ln(\sqrt{3})$ .

**Warrenx** · December 9th 2010, 04:11 AM

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

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