What is the integral of (2x+1)/((x^2+4)^2(x+1)^3) ?

The integral of (2x+1)/((x^2+4)^2(x+1)^3) is

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integral of (2x+1)/((x^2+4)^2(x+1)^3)

\int \frac{2 x + 1}{( x ^{2} + 4 ) ^{2} ( x + 1 ) ^{3}} d x

- \frac{31}{10000} arctan (\frac{x}{2}) - \frac{19}{625} ln x^{2} + 4 + \frac{38}{625} ln ∣ x + 1 ∣ - \frac{19}{4000} sin (2 arctan (\frac{x}{2})) + \frac{21}{250 ( x ^{2} + 4 )} - \frac{6}{125 ( x + 1 )} + \frac{1}{50 ( x + 1 ) ^{2}} + C

Solution steps

\int \frac{2 x + 1}{( x ^{2} + 4 ) ^{2} ( x + 1 ) ^{3}} d x

Take the partial fraction of

\frac{2 x + 1}{( x ^{2} + 4 ) ^{2} ( x + 1 ) ^{3}} : \frac{- 38 x + 8}{625 ( x ^{2} + 4 )} + \frac{- 21 x - 19}{125 ( x ^{2} + 4 ) ^{2}} + \frac{38}{625 ( x + 1 )} + \frac{6}{125 ( x + 1 ) ^{2}} - \frac{1}{25 ( x + 1 ) ^{3}}

= \int \frac{- 38 x + 8}{625 ( x ^{2} + 4 )} + \frac{- 21 x - 19}{125 ( x ^{2} + 4 ) ^{2}} + \frac{38}{625 ( x + 1 )} + \frac{6}{125 ( x + 1 ) ^{2}} - \frac{1}{25 ( x + 1 ) ^{3}} 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{- 38 x + 8}{625 ( x ^{2} + 4 )} d x + \int \frac{- 21 x - 19}{125 ( x ^{2} + 4 ) ^{2}} d x + \int \frac{38}{625 ( x + 1 )} d x + \int \frac{6}{125 ( x + 1 ) ^{2}} d x - \int \frac{1}{25 ( x + 1 ) ^{3}} d x

\int \frac{- 38 x + 8}{625 ( x ^{2} + 4 )} d x = \frac{1}{625} (- 19 ln x^{2} + 4 + 4 arctan (\frac{x}{2}))

\int \frac{- 21 x - 19}{125 ( x ^{2} + 4 ) ^{2}} d x = \frac{1}{125} (\frac{21}{2 ( x ^{2} + 4 )} - \frac{19}{16} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))))

\int \frac{38}{625 ( x + 1 )} d x = \frac{38}{625} ln ∣ x + 1 ∣

\int \frac{6}{125 ( x + 1 ) ^{2}} d x = - \frac{6}{125 ( x + 1 )}

\int \frac{1}{25 ( x + 1 ) ^{3}} d x = - \frac{1}{50 ( x + 1 ) ^{2}}

= \frac{1}{625} (- 19 ln x^{2} + 4 + 4 arctan (\frac{x}{2})) + \frac{1}{125} (\frac{21}{2 ( x ^{2} + 4 )} - \frac{19}{16} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))) + \frac{38}{625} ln ∣ x + 1 ∣ - \frac{6}{125 ( x + 1 )} - (- \frac{1}{50 ( x + 1 ) ^{2}})

Simplify

\frac{1}{625} (- 19 ln x^{2} + 4 + 4 arctan (\frac{x}{2})) + \frac{1}{125} (\frac{21}{2 ( x ^{2} + 4 )} - \frac{19}{16} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))) + \frac{38}{625} ln ∣ x + 1 ∣ - \frac{6}{125 ( x + 1 )} - (- \frac{1}{50 ( x + 1 ) ^{2}}) : - \frac{31}{10000} arctan (\frac{x}{2}) - \frac{19}{625} ln x^{2} + 4 + \frac{38}{625} ln ∣ x + 1 ∣ - \frac{19}{4000} sin (2 arctan (\frac{x}{2})) + \frac{21}{250 ( x ^{2} + 4 )} - \frac{6}{125 ( x + 1 )} + \frac{1}{50 ( x + 1 ) ^{2}}

= - \frac{31}{10000} arctan (\frac{x}{2}) - \frac{19}{625} ln x^{2} + 4 + \frac{38}{625} ln ∣ x + 1 ∣ - \frac{19}{4000} sin (2 arctan (\frac{x}{2})) + \frac{21}{250 ( x ^{2} + 4 )} - \frac{6}{125 ( x + 1 )} + \frac{1}{50 ( x + 1 ) ^{2}}

Add a constant to the solution

= - \frac{31}{10000} arctan (\frac{x}{2}) - \frac{19}{625} ln x^{2} + 4 + \frac{38}{625} ln ∣ x + 1 ∣ - \frac{19}{4000} sin (2 arctan (\frac{x}{2})) + \frac{21}{250 ( x ^{2} + 4 )} - \frac{6}{125 ( x + 1 )} + \frac{1}{50 ( x + 1 ) ^{2}} + C

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What is the integral of (2x+1)/((x^2+4)^2(x+1)^3) ?
The integral of (2x+1)/((x^2+4)^2(x+1)^3) is