`int (x^2 + 1)/(x^2 - 2x + 2)^2 dx` Evaluate the integral

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

Let's evaluate the above integral by rewriting the integrand as,

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

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

`=int1/(x^2-2x+2)dx+int(2x-1)/(x^2-2x+2)^2dx`

Now again rewrite the second integral,

`=int1/(x^2-2x+2)dx+int((2x-2)+1)/(x^2-2x+2)^2dx`

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

Now let's evaluate the above three integrals,

`int1/(x^2-2x+2)dx=int1/((x-1)^2+1)dx`

Let's use the integral substitution,

Let u=x-1,

du=dx

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

The above can be evaluated using the standard integral,

`int1/(x^2+a^2)dx=1/atan^(-1)(x/a)`

`=1/1tan^(-1)(u/1)`

`=tan^(-1)(u)`

Substitute back u=x-1,

`=tan^(-1)(x-1)`

Now let's evaluate the second integral by integral substitution,

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

Let `v=(x^2-2x+2)`

`dv=(2x-2)dx`

`=int1/v^2dv`

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

`=v^(-1)/-1`

`=-1/v`

Substitute back `v=(x^2-2x+2)`

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

Noe let's evaluate the third integral,

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

`=int1/((x-1)^2+1)^2dx`

Let's use the integral substitution,

Let `tan(y)=x-1`

`sec^2(y)dy=dx`

`=int(sec^2(y)dy)/(tan^2(y)+1)^2`

Using the identity:`1+tan^2(y)=sec^2(y)`

`=int(sec^2(y))/(sec^2(y))^2dy`

`=int1/(sec^2(y))dy`

`=intcos^2(y)dy`

Now use the identity:`cos^2(y)=(1+cos(2y))/2`

`=int(1+cos(2y))/2dy`

`=int(1dy)/2+intcos(2y)/2dy`

` ` `=y/2+1/2sin(2y)/2`

`=y/2+sin(2y)/4`

Substitute back `y=tan^(-1)(x-1)`

`=1/2arctan(x-1)+1/4sin(2arctan(x-1))`

`=1/2arctan(x-1)+1/4{2sin(arctan(x-1))cos(arctan(x-1))}`

`=1/2arctan(x-1)+1/4{2*(x-1)/sqrt(x^2-2x+2)*1/sqrt(x^2-2x+2)}`

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

`:.int(x^2+1)/(x^2-2x+2)^2dx=arctan(x-1)-1/(x^2-2x+2)+1/2arctan(x-1)+(x-1)/(2(x^2-2x+2))`

Add a constant C to the solution and simplify,

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

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