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

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

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

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

\frac{1}{9} ln ∣ x + 1 ∣ + \frac{1}{18} (- ln x^{2} + 2 + 2 arctan (\frac{x}{2})) + \frac{1}{3} (\frac{1}{4 2} arctan (\frac{1}{2} x) + \frac{1}{8 2} sin (2 arctan (\frac{1}{2} x)) + \frac{1}{2 ( x ^{2} + 2 )}) + C

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

\frac{1}{u ( u ^{2} - 2 u + 3 ) ^{2}} : \frac{1}{9 u} + \frac{- u + 2}{9 ( u ^{2} - 2 u + 3 )} + \frac{- u + 2}{3 ( u ^{2} - 2 u + 3 ) ^{2}}

= \int \frac{1}{9 u} + \frac{- u + 2}{9 ( u ^{2} - 2 u + 3 )} + \frac{- u + 2}{3 ( u ^{2} - 2 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}{9 u} d u + \int \frac{- u + 2}{9 ( u ^{2} - 2 u + 3 )} d u + \int \frac{- u + 2}{3 ( u ^{2} - 2 u + 3 ) ^{2}} d u

\int \frac{1}{9 u} d u = \frac{1}{9} ln ∣ u ∣

\int \frac{- u + 2}{9 ( u ^{2} - 2 u + 3 )} d u = \frac{1}{9} (- \frac{1}{2} ln u^{2} - 2 u + 3 - \frac{1}{2} arctan (\frac{1}{2} (u - 1)) + 2 arctan (\frac{1}{2} (u - 1)))

\int \frac{- u + 2}{3 ( u ^{2} - 2 u + 3 ) ^{2}} d u = \frac{1}{3} (\frac{1}{4 2} arctan (\frac{1}{2} (u - 1)) + \frac{1}{8 2} sin (2 arctan (\frac{1}{2} (u - 1))) + \frac{1}{2 ( u ^{2} - 2 u + 3 )})

= \frac{1}{9} ln ∣ u ∣ + \frac{1}{9} (- \frac{1}{2} ln u^{2} - 2 u + 3 - \frac{1}{2} arctan (\frac{1}{2} (u - 1)) + 2 arctan (\frac{1}{2} (u - 1))) + \frac{1}{3} (\frac{1}{4 2} arctan (\frac{1}{2} (u - 1)) + \frac{1}{8 2} sin (2 arctan (\frac{1}{2} (u - 1))) + \frac{1}{2 ( u ^{2} - 2 u + 3 )})

Substitute back

u = x + 1

= \frac{1}{9} ln ∣ x + 1 ∣ + \frac{1}{9} (- \frac{1}{2} ln (x + 1)^{2} - 2 (x + 1) + 3 - \frac{1}{2} arctan (\frac{1}{2} (x + 1 - 1)) + 2 arctan (\frac{1}{2} (x + 1 - 1))) + \frac{1}{3} \frac{1}{4 2} arctan (\frac{1}{2} (x + 1 - 1)) + \frac{1}{8 2} sin (2 arctan (\frac{1}{2} (x + 1 - 1))) + \frac{1}{2 ( ( x + 1 ) ^{2} - 2 ( x + 1 ) + 3 )}

Simplify

\frac{1}{9} ln ∣ x + 1 ∣ + \frac{1}{9} (- \frac{1}{2} ln (x + 1)^{2} - 2 (x + 1) + 3 - \frac{1}{2} arctan (\frac{1}{2} (x + 1 - 1)) + 2 arctan (\frac{1}{2} (x + 1 - 1))) + \frac{1}{3} \frac{1}{4 2} arctan (\frac{1}{2} (x + 1 - 1)) + \frac{1}{8 2} sin (2 arctan (\frac{1}{2} (x + 1 - 1))) + \frac{1}{2 ( ( x + 1 ) ^{2} - 2 ( x + 1 ) + 3 )} : \frac{1}{9} ln ∣ x + 1 ∣ + \frac{1}{18} (- ln x^{2} + 2 + 2 arctan (\frac{x}{2})) + \frac{1}{3} (\frac{1}{4 2} arctan (\frac{1}{2} x) + \frac{1}{8 2} sin (2 arctan (\frac{1}{2} x)) + \frac{1}{2 ( x ^{2} + 2 )})

= \frac{1}{9} ln ∣ x + 1 ∣ + \frac{1}{18} (- ln x^{2} + 2 + 2 arctan (\frac{x}{2})) + \frac{1}{3} (\frac{1}{4 2} arctan (\frac{1}{2} x) + \frac{1}{8 2} sin (2 arctan (\frac{1}{2} x)) + \frac{1}{2 ( x ^{2} + 2 )})

Add a constant to the solution

= \frac{1}{9} ln ∣ x + 1 ∣ + \frac{1}{18} (- ln x^{2} + 2 + 2 arctan (\frac{x}{2})) + \frac{1}{3} (\frac{1}{4 2} arctan (\frac{1}{2} x) + \frac{1}{8 2} sin (2 arctan (\frac{1}{2} x)) + \frac{1}{2 ( x ^{2} + 2 )}) + C

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