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

The integral of 1/((x+2)(x^2+2x+4)^2) is

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

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

\frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} (- \frac{1}{2} ln x^{2} + 2 x + 4 + \frac{1}{3} arctan (\frac{1}{3} (x + 1))) + \frac{1}{4} - \frac{1}{12 3} x (2 arctan (\frac{x + 1}{3}) + sin (2 arctan (\frac{x + 1}{3}))) + \frac{1}{6} \frac{( x + 1 ) ( arctan ( \frac{1}{3} ( x + 1 ) ) ( x ^{2} + 2 x + 4 ) + 3 ( x + 1 ) )}{3 ( x ^{2} + 2 x + 4 )} + \frac{3}{x ^{2} + 2 x + 4} + C

Solution steps

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

Complete the square

x^{2} + 2 x + 4 : (x + 1)^{2} + 3

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

Apply u-substitution

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

Take the partial fraction of

\frac{1}{( u + 1 ) ( u ^{2} + 3 ) ^{2}} : \frac{1}{16 ( u + 1 )} + \frac{- u + 1}{16 ( u ^{2} + 3 )} + \frac{- u + 1}{4 ( u ^{2} + 3 ) ^{2}}

= \int \frac{1}{16 ( u + 1 )} + \frac{- u + 1}{16 ( u ^{2} + 3 )} + \frac{- u + 1}{4 ( u ^{2} + 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}{16 ( u + 1 )} d u + \int \frac{- u + 1}{16 ( u ^{2} + 3 )} d u + \int \frac{- u + 1}{4 ( u ^{2} + 3 ) ^{2}} d u

\int \frac{1}{16 ( u + 1 )} d u = \frac{1}{16} ln ∣ u + 1 ∣

\int \frac{- u + 1}{16 ( u ^{2} + 3 )} d u = \frac{1}{16} (- \frac{1}{2} ln u^{2} + 3 + \frac{1}{3} arctan (\frac{u}{3}))

\int \frac{- u + 1}{4 ( u ^{2} + 3 ) ^{2}} d u = \frac{1}{4} \frac{1}{6 3} (- u + 1) (arctan (\frac{u}{3}) + \frac{1}{2} sin (2 arctan (\frac{u}{3}))) + \frac{1}{6} \frac{u ( arctan ( \frac{u}{3} ) ( 3 + u ^{2} ) + 3 u )}{3 ( 3 + u ^{2} )} + \frac{3}{u ^{2} + 3}

= \frac{1}{16} ln ∣ u + 1 ∣ + \frac{1}{16} (- \frac{1}{2} ln u^{2} + 3 + \frac{1}{3} arctan (\frac{u}{3})) + \frac{1}{4} \frac{1}{6 3} (- u + 1) (arctan (\frac{u}{3}) + \frac{1}{2} sin (2 arctan (\frac{u}{3}))) + \frac{1}{6} \frac{u ( arctan ( \frac{u}{3} ) ( 3 + u ^{2} ) + 3 u )}{3 ( 3 + u ^{2} )} + \frac{3}{u ^{2} + 3}

Substitute back

u = x + 1

= \frac{1}{16} ln ∣ x + 1 + 1 ∣ + \frac{1}{16} (- \frac{1}{2} ln (x + 1)^{2} + 3 + \frac{1}{3} arctan (\frac{x + 1}{3})) + \frac{1}{4} \frac{1}{6 3} (- (x + 1) + 1) (arctan (\frac{x + 1}{3}) + \frac{1}{2} sin (2 arctan (\frac{x + 1}{3}))) + \frac{1}{6} \frac{( x + 1 ) ( arctan ( \frac{x + 1}{3} ) ( 3 + ( x + 1 ) ^{2} ) + 3 ( x + 1 ) )}{3 ( 3 + ( x + 1 ) ^{2} )} + \frac{3}{( x + 1 ) ^{2} + 3}

Simplify

\frac{1}{16} ln ∣ x + 1 + 1 ∣ + \frac{1}{16} (- \frac{1}{2} ln (x + 1)^{2} + 3 + \frac{1}{3} arctan (\frac{x + 1}{3})) + \frac{1}{4} \frac{1}{6 3} (- (x + 1) + 1) (arctan (\frac{x + 1}{3}) + \frac{1}{2} sin (2 arctan (\frac{x + 1}{3}))) + \frac{1}{6} \frac{( x + 1 ) ( arctan ( \frac{x + 1}{3} ) ( 3 + ( x + 1 ) ^{2} ) + 3 ( x + 1 ) )}{3 ( 3 + ( x + 1 ) ^{2} )} + \frac{3}{( x + 1 ) ^{2} + 3} : \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} (- \frac{1}{2} ln x^{2} + 2 x + 4 + \frac{1}{3} arctan (\frac{1}{3} (x + 1))) + \frac{1}{4} - \frac{1}{12 3} x (2 arctan (\frac{x + 1}{3}) + sin (2 arctan (\frac{x + 1}{3}))) + \frac{1}{6} \frac{( x + 1 ) ( arctan ( \frac{1}{3} ( x + 1 ) ) ( x ^{2} + 2 x + 4 ) + 3 ( x + 1 ) )}{3 ( x ^{2} + 2 x + 4 )} + \frac{3}{x ^{2} + 2 x + 4}

= \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} (- \frac{1}{2} ln x^{2} + 2 x + 4 + \frac{1}{3} arctan (\frac{1}{3} (x + 1))) + \frac{1}{4} - \frac{1}{12 3} x (2 arctan (\frac{x + 1}{3}) + sin (2 arctan (\frac{x + 1}{3}))) + \frac{1}{6} \frac{( x + 1 ) ( arctan ( \frac{1}{3} ( x + 1 ) ) ( x ^{2} + 2 x + 4 ) + 3 ( x + 1 ) )}{3 ( x ^{2} + 2 x + 4 )} + \frac{3}{x ^{2} + 2 x + 4}

Add a constant to the solution

= \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} (- \frac{1}{2} ln x^{2} + 2 x + 4 + \frac{1}{3} arctan (\frac{1}{3} (x + 1))) + \frac{1}{4} - \frac{1}{12 3} x (2 arctan (\frac{x + 1}{3}) + sin (2 arctan (\frac{x + 1}{3}))) + \frac{1}{6} \frac{( x + 1 ) ( arctan ( \frac{1}{3} ( x + 1 ) ) ( x ^{2} + 2 x + 4 ) + 3 ( x + 1 ) )}{3 ( x ^{2} + 2 x + 4 )} + \frac{3}{x ^{2} + 2 x + 4} + C

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