# `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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