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#