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