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

*print*Print*list*Cite

### 2 Answers

`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

`=intsqrt(13)cosydy/sqrt(13-13sin^2y)`

`=int{sqrt(13)cosy}/{sqrt(13).sqrt(1-sin^2y)}dy`

`=intcosy/cosydy`

`=int1dy`

`=y`

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

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`