What is $int 23/sqrt(13+3x^2) dx$ ?

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Answer 1 · 2015-11-12T13:01:59

$23int1/sqrt(13+3x^2)dx$

$x = sqrt(13/3)u$

$dx = sqrt(13/3)du$

$u = sqrt(3/13)x$

$23sqrt(13/3)int 1/sqrt(13+13u^2)du$

$23sqrt(13/3)*sqrt(1/13)int 1/sqrt(1+u^2)du$

$23sqrt(1/3)[arcsinh(u)]+C$

Substitute back

$23sqrt(1/3)[arcsinh(sqrt(3/13)x)]+C$

What is int 23/sqrt(13+3x^2) dxint 23/sqrt(13+3x^2) dx?