`int x/(x^2-6x+10)^2 dx` Use integration tables to find the indefinite integral.

`intx/(x^2-6x+10)^2dx`

Let's rewrite the integrand as,

`=1/2int(2x)/(x^2-6x+10)^2dx`

`=1/2int(2x-6+6)/(x^2-6x+10)^2dx`

`=1/2[int(2x-6)/(x^2-6x+10)^2dx+int6/(x^2-6x+10)^2dx]`  --------------------(1)

Now let' evaluate each of the above two integrals separately,

`int(2x-6)/(x^2-6x+10)^2dx`

Let's apply integral substitution:`u=x^2-6x+10`

`=>du=(2x-6)dx`

`=int1/u^2du`

`=intu^(-2)du`

Now from the integer tables:`intu^ndu=u^(n+1)/(n+1)+C`

`=u^(-2+1)/(-2+1)`

`=-1/u`

Substitute back `u=x^2-6x+10`

`=-1/(x^2-6x+10)`                    -----------------------------(2)

Now let's evaluate the second integral,

`int6/(x^2-6x+10)^2dx`

Take the constant out,

`=6int1/(x^2-6x+10)^2dx`

Complete the square of the term in the denominator.

`=6int1/((x-3)^2+1)^2dx`  

Let's apply integral substitution:`u=x-3`

...

`intx/(x^2-6x+10)^2dx`

Let's rewrite the integrand as,

`=1/2int(2x)/(x^2-6x+10)^2dx`

`=1/2int(2x-6+6)/(x^2-6x+10)^2dx`

`=1/2[int(2x-6)/(x^2-6x+10)^2dx+int6/(x^2-6x+10)^2dx]`  --------------------(1)

Now let' evaluate each of the above two integrals separately,

`int(2x-6)/(x^2-6x+10)^2dx`

Let's apply integral substitution:`u=x^2-6x+10`

`=>du=(2x-6)dx`

`=int1/u^2du`

`=intu^(-2)du`

Now from the integer tables:`intu^ndu=u^(n+1)/(n+1)+C`

`=u^(-2+1)/(-2+1)`

`=-1/u`

Substitute back `u=x^2-6x+10`

`=-1/(x^2-6x+10)`                    -----------------------------(2)

Now let's evaluate the second integral,

`int6/(x^2-6x+10)^2dx`

Take the constant out,

`=6int1/(x^2-6x+10)^2dx`

Complete the square of the term in the denominator.

`=6int1/((x-3)^2+1)^2dx`  

Let's apply integral substitution:`u=x-3`

`=>du=dx`

`=6int1/(u^2+1^2)^2du`

Now use the following from the integration tables:

`int1/(a^2+-u^2)^ndu=1/(2a^2(n-1))[u/(a^2+-u^2)^(n-1)+(2n-3)int1/(a^2+-u^2)^(n-1)du]`

`=6{1/(2(1)^2(2-1))[u/(1^2+u^2)^(2-1)+(2(2)-3)int1/(1^2+u^2)^(2-1)du]}`

`=6{1/2[u/(1+u^2)+int1/(1^2+u^2)du]}`

Now from the integration table:`int1/(a^2+u^2)du=1/aarctan(u/a)+C`

`=6{1/2[u/(1+u^2)+arctan(u/1)]}`

`=(3u)/(1+u^2)+3arctan(u)`

Substitute back `u=x-3`

`=(3(x-3))/(1+(x-3)^2)+3arctan(x-3)`

`=(3x-9)/(1+x^2-6x+9)+3arctan(x-3)`

`=(3x-9)/(x^2-6x+10)+3arctan(x-3)`    -------------------------(3)

Plug back the results of the integrals 2 and 3 in 1

`int1/(x^2-6x+10)^2dx=1/2[-1/(x^2-6x+10)+(3x-9)/(x^2-6x+10)+3arctan(x-3)]`

`=1/2[(3x-9-1)/(x^2-6x+10)+3arctan(x-3)]`

`=1/2[(3x-10)/(x^2-6x+10)+3arctan(x-3)]`

`=(3x-10)/(2(x^2-6x+10))+3/2arctan(x-3)`

Add a constant C to the solution,

`=(3x-10)/(2(x^2-6x+10))+3/2arctan(x-3)+C`