What is the integral of 1/((x^2-4x+4)(x^2-4x+5)) ?

The integral of 1/((x^2-4x+4)(x^2-4x+5)) is -1/(x-2)-arctan(x-2)+C

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integral of 1/((x^2-4x+4)(x^2-4x+5))

\int \frac{1}{( x ^{2} - 4 x + 4 ) ( x ^{2} - 4 x + 5 )} d x

- \frac{1}{x - 2} - arctan (x - 2) + C

Solution steps

\int \frac{1}{( x ^{2} - 4 x + 4 ) ( x ^{2} - 4 x + 5 )} d x

Complete the square

x^{2} - 4 x + 4 : (x - 2)^{2}

= \int \frac{1}{( x - 2 ) ^{2} ( x ^{2} - 4 x + 5 )} d x

Apply u-substitution

: \int \frac{1}{u ^{2} ( u ^{2} + 1 )} d u

= \int \frac{1}{u ^{2} ( u ^{2} + 1 )} d u

Take the partial fraction of

\frac{1}{u ^{2} ( u ^{2} + 1 )} : \frac{1}{u ^{2}} - \frac{1}{u ^{2} + 1}

= \int \frac{1}{u ^{2}} - \frac{1}{u ^{2} + 1} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{u ^{2}} d u - \int \frac{1}{u ^{2} + 1} d u

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= - \frac{1}{u} - arctan (u)

Substitute back

u = x - 2

= - \frac{1}{x - 2} - arctan (x - 2)

Add a constant to the solution

= - \frac{1}{x - 2} - arctan (x - 2) + C

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What is the integral of 1/((x^2-4x+4)(x^2-4x+5)) ?
The integral of 1/((x^2-4x+4)(x^2-4x+5)) is -1/(x-2)-arctan(x-2)+C