Integral of 1/sqrt(x^2-9/4): why is it log(2x+sqrt(4x^2-9))?

[green_tea]

Hi!
Can anyone explain to me why the integrale of 1/sqrt(x^2-9/4) is log(2x + sqrt(4x^2 -9)) ?

I know this formula: the integrale of 1/sqrt(1+x^2) = log (absolute value of: x + sqrt(1+x^2)) where log is the natural logarithm. Can I use that to solve my problem? If I can, than how?

 

[royhaas]

Re: Integrate 1/sqrt(x^2-9/4)

Use \(\displaystyle x=(3/2)\sec(\theta)\).

 

[qpmathelp]

Re: Integrate 1/sqrt(x^2-9/4)

you can use the substitution in the second post or

use the direct formula for integral of

\(\displaystyle {1 \over {\sqrt {x^2 - a^2 } }}\)

try the following link for some formulae
http://rbmix.com/math/calculus/integration/int.php