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

Let's complete the square for the denominator of the integral as:

`(x^2-4x+9)=(x-2)^2+5`

`(x-2)^2+(sqrt(5))^2`

`int1/(x^2-4x+9)dx=int1/((x-2)^2+(sqrt(5))^2)dx`

Let's apply the integral substitution,

substitute `u=x-2`

`du=1dx`

`=int1/(u^2+(sqrt(5))^2)du`

Now use the standard integral :`int1/(x^2+a^2)=1/aarctan(x/a)`

`=1/sqrt(5)arctan(u/sqrt(5))`

substitute back u=(x-2) and add a constant C to the solution,

`=1/sqrt(5)arctan((x-2)/sqrt(5))+C`

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

Let's complete the square for the denominator of the integral as:

`(x^2-4x+9)=(x-2)^2+5`

`(x-2)^2+(sqrt(5))^2`

`int1/(x^2-4x+9)dx=int1/((x-2)^2+(sqrt(5))^2)dx`

Let's apply the integral substitution,

substitute `u=x-2`

`du=1dx`

`=int1/(u^2+(sqrt(5))^2)du`

Now use the standard integral :`int1/(x^2+a^2)=1/aarctan(x/a)`

`=1/sqrt(5)arctan(u/sqrt(5))`

substitute back u=(x-2) and add a constant C to the solution,

`=1/sqrt(5)arctan((x-2)/sqrt(5))+C`