How do you integrate $1/(sqrt(25+4x^2))$ ?

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Answer 1 · 2015-02-01T19:09:56

I would take out, from the square root, $25$ getting:

$int1/(5*sqrt(1+4/25x^2))dx=$
$=int1/5*1/(sqrt(1+(2/5x)^2))dx=$
$=1/5*5/2sinh^(-1)(2/5x)+c$
$=1/2sinh^(-1)(2/5x)+c$

Where $sinh^(-1)(x)=$ arcsinh(x) is one of the inverse hyperbolic functions .
Have a look at: http://mathworld.wolfram.com/InverseHyperbolicSine.html
for more info.

How do you integrate 1/(sqrt(25+4x^2))1/(sqrt(25+4x^2)) ?