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Evaluate the integral of 1 / ((x2 – 4x + 4)(x2 – 4x + 5))

Compute the following integral.

    \[ \int \frac{dx}{(x^2 - 4x + 4)(x^2 - 4x + 5)}. \]


In the denominator we have

    \[ (x^2 - 4x + 4)(x^2 - 4x + 5) = (x-2)^2 (x^2 - 4x + 5).  \]

Then we use partial fractions,

    \[ \frac{1}{(x-2)^2 (x^2 - 4x + 5)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{Cx + D}{x^2 - 4x + 5}. \]

This gives us the equation

    \[ A(x-2)(x^2 - 4x + 5) + B(x^2 - 4x + 5) + (Cx + D)(x-2)^2 = 1. \]

We evaluate at x = 2 to obtain a value for B,

    \[ B = 1. \]

Then using this value of B and evaluating at x = -1, x = 0 and x = 1 to obtain

    \begin{align*}  -30A + 10 - 9C + 9D &= 1 \\  -10A + 5 + 4D &= 1 \\  -2A + 2 + C + D &= 1. \end{align*}

Solving this system of equations we obtain

    \[ A = 0, \qquad C = 0, \qquad D = -1. \]

Therefore, we have

    \begin{align*}  \int \frac{dx}{(x^2 - 4x + 4)(x^2 - 4x + 5)} &= \int \left( \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{Cx+D}{x^2 - 4x + 5} \right) \, dx \\[10pt]  &= \int \frac{1}{(x-2)^2} \, dx - \int \frac{1}{x^2 - 4x + 5} \, dx \\[10pt]  &= \frac{-1}{x-2} - \int \frac{1}{(x-2)^2 + 1} \, dx \\[10pt]  &= \frac{-1}{x-2} - \arctan (x-2) + C. \end{align*}

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