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

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

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

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

- \frac{1}{160} x sin (2 arctan (\frac{x}{2})) + \frac{1}{25} ln ∣ x + 1 ∣ - \frac{1}{50} ln x^{2} + 4 + \frac{13}{400} arctan (\frac{x}{2}) + \frac{1}{160} sin (2 arctan (\frac{x}{2})) + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x + 1 ) ( x ^{2} + 4 ) ^{2}} : \frac{1}{25 ( x + 1 )} + \frac{- x + 1}{25 ( x ^{2} + 4 )} + \frac{- x + 1}{5 ( x ^{2} + 4 ) ^{2}}

= \int \frac{1}{25 ( x + 1 )} + \frac{- x + 1}{25 ( x ^{2} + 4 )} + \frac{- x + 1}{5 ( x ^{2} + 4 ) ^{2}} d x

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}{25 ( x + 1 )} d x + \int \frac{- x + 1}{25 ( x ^{2} + 4 )} d x + \int \frac{- x + 1}{5 ( x ^{2} + 4 ) ^{2}} d x

\int \frac{1}{25 ( x + 1 )} d x = \frac{1}{25} ln ∣ x + 1 ∣

\int \frac{- x + 1}{25 ( x ^{2} + 4 )} d x = \frac{1}{25} (- \frac{1}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}))

\int \frac{- x + 1}{5 ( x ^{2} + 4 ) ^{2}} d x = \frac{1}{160} (- x sin (2 arctan (\frac{x}{2})) + 2 arctan (\frac{x}{2}) + sin (2 arctan (\frac{x}{2})))

= \frac{1}{25} ln ∣ x + 1 ∣ + \frac{1}{25} (- \frac{1}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2})) + \frac{1}{160} (- x sin (2 arctan (\frac{x}{2})) + 2 arctan (\frac{x}{2}) + sin (2 arctan (\frac{x}{2})))

Simplify

\frac{1}{25} ln ∣ x + 1 ∣ + \frac{1}{25} (- \frac{1}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2})) + \frac{1}{160} (- x sin (2 arctan (\frac{x}{2})) + 2 arctan (\frac{x}{2}) + sin (2 arctan (\frac{x}{2}))) : - \frac{1}{160} x sin (2 arctan (\frac{x}{2})) + \frac{1}{25} ln ∣ x + 1 ∣ - \frac{1}{50} ln x^{2} + 4 + \frac{13}{400} arctan (\frac{x}{2}) + \frac{1}{160} sin (2 arctan (\frac{x}{2}))

= - \frac{1}{160} x sin (2 arctan (\frac{x}{2})) + \frac{1}{25} ln ∣ x + 1 ∣ - \frac{1}{50} ln x^{2} + 4 + \frac{13}{400} arctan (\frac{x}{2}) + \frac{1}{160} sin (2 arctan (\frac{x}{2}))

Add a constant to the solution

= - \frac{1}{160} x sin (2 arctan (\frac{x}{2})) + \frac{1}{25} ln ∣ x + 1 ∣ - \frac{1}{50} ln x^{2} + 4 + \frac{13}{400} arctan (\frac{x}{2}) + \frac{1}{160} sin (2 arctan (\frac{x}{2})) + C

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