How do you integrate $int 1/(x^2sqrt(4+x^2))$ by trigonometric substitution?

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Answer 1 · 2018-04-08T12:01:39

The answer is $=-sqrt(4+x^2)/(4x)+C$

Perform this integral by substitution

Let $x=2tanu$ , $=>$ , $dx=2sec^2udu$

$sqrt(4+x^2)=sqrt(4+4tan^2u)=2secu$

The integral is

$I=int(2sec^2udu)/(4tan^2u*2secu)$

$=1/4int(sec udu)/(tan^2u)$

$=1/4int(cosu du)/(sin^2u)$

Let $v=sinu$ , $=>$ , $dv=cosu du$

Therefore,

$I=1/4int(dv)/(v^2)$

$=1/4*-(1/v)$

$=-1/(4sinu)$

$=-1/(4*x/(sqrt(x^2+4)))$

$=-sqrt(4+x^2)/(4x)+C$

How do you integrate int 1/(x^2sqrt(4+x^2))int 1/(x^2sqrt(4+x^2)) by trigonometric substitution?