What is the integral of 1/sqrt(9-x^2+4)

Expert Answers
justaguide eNotes educator| Certified Educator

The integral `int 1/sqrt(9-x^2+4) dx` has to be determined.

`int 1/sqrt(9-x^2+4) dx`

=> `int 1/sqrt(13 - x^2) dx`

=> `int 1/((sqrt 13)*sqrt(1 - (x/sqrt13)^2)) dx`

let `x/sqrt 13 = y => sqrt 13*dy = dx`

=> `int sqrt 13/(sqrt 13)*sqrt(1 - y^2) dy`

=> `int 1/sqrt(1 - y^2) dy`

=> `sin^-1y`

=> `sin^-1(x/sqrt 13)`

The integral `int 1/sqrt(9-x^2+4) dx = sin^-1(x/sqrt 13) + C`

rakesh05 eNotes educator| Certified Educator

`int1/sqrt(9-x^2+4)dx` =`int1/sqrt(13-x^2)dx`

Take `x=sqrt(13)siny`    or  `y=sin^-1(x/sqrt(13))`

     so `dx=sqrt(13)cosydy`

So the the above integral expression can be written as






`=sin^-1(x/sqrt(13))` .        Answer.