What is the integral of 1/((x^2+6x+6)^2) ?

The integral of 1/((x^2+6x+6)^2) is

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integral of 1/((x^2+6x+6)^2)

\int \frac{1}{( x ^{2} + 6 x + 6 ) ^{2}} d x

\frac{1}{12 3} ln x + 3 + 3 - \frac{1}{12 ( x + 3 + 3 )} - \frac{1}{12 3} ln x + 3 - 3 - \frac{1}{12 ( x + 3 - 3 )} + C

Solution steps

\int \frac{1}{( x ^{2} + 6 x + 6 ) ^{2}} d x

Complete the square

x^{2} + 6 x + 6 : (x + 3)^{2} - 3

= \int \frac{1}{( ( x + 3 ) ^{2} - 3 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{( u ^{2} - 3 ) ^{2}} d u

= \int \frac{1}{( u ^{2} - 3 ) ^{2}} d u

Take the partial fraction of

\frac{1}{( u ^{2} - 3 ) ^{2}} : \frac{1}{12 3 ( u + 3 )} + \frac{1}{12 ( u + 3 ) ^{2}} - \frac{1}{12 3 ( u - 3 )} + \frac{1}{12 ( u - 3 ) ^{2}}

= \int \frac{1}{12 3 ( u + 3 )} + \frac{1}{12 ( u + 3 ) ^{2}} - \frac{1}{12 3 ( u - 3 )} + \frac{1}{12 ( u - 3 ) ^{2}} 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}{12 3 ( u + 3 )} d u + \int \frac{1}{12 ( u + 3 ) ^{2}} d u - \int \frac{1}{12 3 ( u - 3 )} d u + \int \frac{1}{12 ( u - 3 ) ^{2}} d u

\int \frac{1}{12 3 ( u + 3 )} d u = \frac{1}{12 3} ln u + 3

\int \frac{1}{12 ( u + 3 ) ^{2}} d u = - \frac{1}{12 ( u + 3 )}

\int \frac{1}{12 3 ( u - 3 )} d u = \frac{1}{12 3} ln u - 3

\int \frac{1}{12 ( u - 3 ) ^{2}} d u = - \frac{1}{12 ( u - 3 )}

= \frac{1}{12 3} ln u + 3 - \frac{1}{12 ( u + 3 )} - \frac{1}{12 3} ln u - 3 - \frac{1}{12 ( u - 3 )}

Substitute back

u = x + 3

= \frac{1}{12 3} ln x + 3 + 3 - \frac{1}{12 ( x + 3 + 3 )} - \frac{1}{12 3} ln x + 3 - 3 - \frac{1}{12 ( x + 3 - 3 )}

Add a constant to the solution

= \frac{1}{12 3} ln x + 3 + 3 - \frac{1}{12 ( x + 3 + 3 )} - \frac{1}{12 3} ln x + 3 - 3 - \frac{1}{12 ( x + 3 - 3 )} + C

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